💡Okay, so I have this problem about an ellipse, and I need to figure out its equation based on some given conditions. Let me try to break it down step by step.First, the ellipse is given by the equation (frac{x^2}{25} + frac{y^2}{b^2} = 1), where (0 < b < 5). I know that for an ellipse, the standard form is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. Since (a^2 = 25), that means (a = 5). Also, since (0 < b < 5), the major axis is along the x-axis.Now, the problem states that the lengths of the major axis, minor axis, and the focal length form an arithmetic sequence. Hmm, let me recall what an arithmetic sequence is. It's a sequence where the difference between consecutive terms is constant. So, if I have three terms (t_1), (t_2), (t_3), then (t_2 - t_1 = t_3 - t_2), which implies (2t_2 = t_1 + t_3).In this case, the three terms are the lengths of the major axis, minor axis, and focal length. Let me denote them:- Length of the major axis: Since the semi-major axis is (a = 5), the major axis length is (2a = 10).- Length of the minor axis: Similarly, the semi-minor axis is (b), so the minor axis length is (2b).- Focal length: The distance between the two foci of the ellipse is (2c), where (c) is the distance from the center to each focus. For an ellipse, (c^2 = a^2 - b^2).So, the three lengths are (10) (major axis), (2b) (minor axis), and (2c) (focal length). These three form an arithmetic sequence. That means:Either (10), (2b), (2c) are in arithmetic sequence, or some other order. Wait, the problem doesn't specify the order, so I need to figure out the correct order.But let's think about it. The major axis is the longest, then the minor axis is shorter, and the focal length is even shorter because (c < a) and (c < b) (since (c^2 = a^2 - b^2), so (c = sqrt{a^2 - b^2}), which is less than (a) and less than (b) because (b < a)).Wait, no. Actually, (c) is the distance from the center to the focus, so (c) is less than (a), but since (b < a), (c) could be greater or less than (b) depending on the values. Let me compute (c) in terms of (b):Given (c^2 = a^2 - b^2 = 25 - b^2), so (c = sqrt{25 - b^2}). Since (b < 5), (c) is positive. Now, is (c) greater or less than (b)?Let me test with (b = 3), then (c = sqrt{25 - 9} = sqrt{16} = 4), which is greater than (b = 3). If (b = 4), then (c = sqrt{25 - 16} = 3), which is less than (b = 4). So, depending on the value of (b), (c) can be greater or less than (b).Hmm, so the order of the lengths isn't immediately clear. Maybe I should consider both possibilities.But let's see: the lengths are 10 (major), 2b (minor), and 2c (focal). Since 10 is the largest, the next could be either 2b or 2c, and the smallest would be the other.But let's think about the arithmetic sequence. If 10, 2b, 2c are in arithmetic sequence, then the difference between 10 and 2b should be equal to the difference between 2b and 2c. So:(2b - 10 = 2c - 2b)Simplify:(2b - 10 = 2c - 2b)Bring like terms together:(2b + 2b - 10 = 2c)(4b - 10 = 2c)Divide both sides by 2:(2b - 5 = c)But we also know that (c = sqrt{25 - b^2}). So:(2b - 5 = sqrt{25 - b^2})Now, let's solve this equation for (b). Let me square both sides to eliminate the square root:((2b - 5)^2 = ( sqrt{25 - b^2} )^2)Expand the left side:(4b^2 - 20b + 25 = 25 - b^2)Bring all terms to one side:(4b^2 - 20b + 25 - 25 + b^2 = 0)Simplify:(5b^2 - 20b = 0)Factor:(5b(b - 4) = 0)So, (b = 0) or (b = 4). But (b > 0), so (b = 4).Wait, but let me check if this is valid. If (b = 4), then (c = 2b - 5 = 8 - 5 = 3). Then, (c = 3), and (c^2 = 9). Let's verify (c^2 = 25 - b^2):(25 - 16 = 9), which is correct. So, (b = 4) is a valid solution.But let me check the other possibility: what if the sequence is 2c, 2b, 10? Then, the difference would be (2b - 2c = 10 - 2b). Let's see:(2b - 2c = 10 - 2b)Simplify:(4b - 2c = 10)Divide by 2:(2b - c = 5)But we also have (c = sqrt{25 - b^2}). So:(2b - sqrt{25 - b^2} = 5)Let me solve this equation:(2b - 5 = sqrt{25 - b^2})Square both sides:((2b - 5)^2 = 25 - b^2)Which is the same equation as before, leading to (b = 4). So, regardless of the order, we get the same solution.Wait, but in the first case, the sequence was 10, 2b, 2c, and in the second case, it's 2c, 2b, 10. But with (b = 4), let's compute the lengths:- Major axis: 10- Minor axis: 2b = 8- Focal length: 2c = 6So, the lengths are 10, 8, 6. Is this an arithmetic sequence?Check the differences: 8 - 10 = -2, 6 - 8 = -2. So, yes, it's an arithmetic sequence with a common difference of -2.Alternatively, if we consider the sequence as 6, 8, 10, which is increasing by 2 each time, that's also an arithmetic sequence. So, depending on the order, both are valid.But in the problem statement, it just says they form an arithmetic sequence, without specifying the order. So, both possibilities are covered, but in both cases, we end up with (b = 4).Therefore, the equation of the ellipse is (frac{x^2}{25} + frac{y^2}{16} = 1), which corresponds to option C.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Identified major axis length as 10, minor as 2b, focal as 2c.2. Set up the arithmetic sequence condition: either 10, 2b, 2c or 2c, 2b, 10.3. Derived equations in both cases, leading to (2b - 5 = c) or (2b - c = 5), both leading to the same quadratic equation.4. Solved for (b = 4), checked validity, and confirmed that the lengths form an arithmetic sequence.5. Concluded that the equation is (frac{x^2}{25} + frac{y^2}{16} = 1), which is option C.Yes, that seems correct. I think I've covered all the necessary steps and checked my work for errors. So, I'm confident that option C is the right answer.

💡First, I notice that the equation has terms with (2^x) on both sides. To simplify, I'll move all (2^x) terms to one side and the constants to the other side.Starting with the original equation:[2^x + 10 = 3 cdot 2^x - 14]I'll subtract (2^x) from both sides to gather the (2^x) terms on the right:[10 = 2 cdot 2^x - 14]Next, I'll add 14 to both sides to isolate the term with (2^x):[24 = 2 cdot 2^x]Now, I'll divide both sides by 2 to solve for (2^x):[12 = 2^x]To find (x), I'll take the logarithm base 2 of both sides:[x = log_2(12)]Since (12) can be expressed as (2^2 cdot 3), I can simplify the logarithm:[x = log_2(2^2 cdot 3) = 2 + log_2(3)]However, the exact form (x = log_2(12)) is sufficient for the solution.

💡Okay, so I have this function f(x) = sin x - x cos x, and I need to do two things with it. First, when x is greater than 0, I need to find the intervals where the function is increasing or decreasing. Second, on the interval from 0 to 2013π, I need to find the sum of all the extreme values. Hmm, let's take this step by step.Starting with part (I): finding the intervals of monotonicity. I remember that to find where a function is increasing or decreasing, I need to look at its first derivative. So, let me find f'(x).Given f(x) = sin x - x cos x. To find f'(x), I'll differentiate term by term.The derivative of sin x is cos x. For the second term, -x cos x, I need to use the product rule. The product rule states that the derivative of u*v is u'v + uv'. So here, u = x and v = cos x. Therefore, u' = 1 and v' = -sin x.Putting it all together, the derivative of -x cos x is -[u'v + uv'] = -[1*cos x + x*(-sin x)] = -cos x + x sin x.So, combining both derivatives, f'(x) = cos x - cos x + x sin x. Wait, that simplifies to f'(x) = x sin x. Okay, that seems right.So, f'(x) = x sin x. Now, to find where the function is increasing or decreasing, I need to determine where f'(x) is positive or negative. Since x > 0, the sign of f'(x) depends on sin x.I know that sin x is positive in the intervals (2kπ, (2k+1)π) and negative in ((2k+1)π, 2(k+1)π) for any integer k. Since x is positive, k will be 0, 1, 2, and so on.Therefore, when sin x is positive, f'(x) is positive, so the function is increasing. When sin x is negative, f'(x) is negative, so the function is decreasing.So, for each k, the function f(x) is increasing on (2kπ, (2k+1)π) and decreasing on ((2k+1)π, 2(k+1)π). That should answer part (I).Moving on to part (II): finding the sum of all extreme values on [0, 2013π]. Extreme values occur where the derivative is zero, which is where f'(x) = x sin x = 0. Since x > 0, sin x = 0. So, sin x = 0 when x is an integer multiple of π, that is, x = kπ for k = 0, 1, 2, ..., 2013.But wait, x is in [0, 2013π], so the critical points are at x = 0, π, 2π, 3π, ..., 2013π. However, x = 0 is an endpoint, so we need to check if it's a maximum or minimum. But since we're considering extreme values, which include both maxima and minima, we need to evaluate f(x) at each critical point and sum them up.But before that, let me confirm if each critical point is a maximum or a minimum. Since the function changes from increasing to decreasing or vice versa at each critical point, we can use the second derivative test or analyze the sign changes of f'(x). Let's see.From part (I), we know that f'(x) is positive on (2kπ, (2k+1)π) and negative on ((2k+1)π, 2(k+1)π). So, at x = (2k+1)π, the derivative changes from positive to negative, indicating a local maximum. At x = 2kπ, the derivative changes from negative to positive, indicating a local minimum.Wait, hold on. Let me think. If k starts at 0, then at x = 0, which is 2*0π, the derivative f'(x) = x sin x. At x = 0, f'(0) = 0. But just to the right of 0, sin x is positive, so f'(x) is positive. So, the function is increasing at x = 0. So, x = 0 is a local minimum? Wait, but f'(x) is increasing from 0 to positive, so it's a minimum. Hmm, but x = 0 is an endpoint, so maybe it's just a minimum.Similarly, at x = π, which is (2*0 +1)π, the derivative changes from positive to negative, so that's a local maximum. At x = 2π, which is 2*1π, the derivative changes from negative to positive, so that's a local minimum, and so on.Therefore, in the interval [0, 2013π], the extreme values occur at x = 0, π, 2π, 3π, ..., 2013π. Each odd multiple of π (i.e., π, 3π, 5π, ...) is a local maximum, and each even multiple of π (i.e., 0, 2π, 4π, ...) is a local minimum.But wait, x = 2013π is the endpoint. Let me check if 2013 is odd or even. 2013 divided by 2 is 1006.5, so it's odd. So, x = 2013π is a local maximum.Therefore, the extreme values are at x = 0 (minimum), x = π (maximum), x = 2π (minimum), x = 3π (maximum), ..., x = 2013π (maximum).So, to find the sum of all extreme values, I need to compute f(0) + f(π) + f(2π) + f(3π) + ... + f(2013π).Let me compute f(kπ) for k = 0, 1, 2, ..., 2013.Given f(x) = sin x - x cos x.At x = kπ, sin(kπ) = 0, because sin(nπ) = 0 for integer n. Cos(kπ) = (-1)^k.Therefore, f(kπ) = 0 - kπ * (-1)^k = kπ * (-1)^{k+1}.So, f(kπ) = (-1)^{k+1} * kπ.Therefore, f(0) = (-1)^{0+1} * 0π = 0.f(π) = (-1)^{1+1} * 1π = 1π.f(2π) = (-1)^{2+1} * 2π = -2π.f(3π) = (-1)^{3+1} * 3π = 3π.f(4π) = (-1)^{4+1} * 4π = -4π.And so on, up to f(2013π) = (-1)^{2013+1} * 2013π = (-1)^{2014} * 2013π = 2013π, since 2014 is even.So, the sum S = f(0) + f(π) + f(2π) + ... + f(2013π) = 0 + π - 2π + 3π - 4π + ... + 2013π.This is an alternating series: 0 + (π - 2π) + (3π - 4π) + ... + (2011π - 2012π) + 2013π.Wait, let's see: the terms are 0, π, -2π, 3π, -4π, ..., 2013π.So, grouping them:(π - 2π) + (3π - 4π) + ... + (2011π - 2012π) + 2013π.Each pair (π - 2π) = -π, (3π - 4π) = -π, and so on.How many such pairs are there?From k=1 to k=2012, stepping by 2, so total number of terms is 2013 (from 0 to 2013π). Excluding f(0)=0, we have 2013 terms, but since we're pairing them, let's see:Wait, actually, the number of terms from k=1 to k=2013 is 2013 terms. Since we're pairing them as (1π - 2π), (3π - 4π), etc., each pair consists of two terms, so the number of pairs is floor(2013/2) = 1006 pairs, and one leftover term, which is 2013π.Wait, 2013 is odd, so 2013 = 2*1006 +1. So, 1006 pairs and one single term.Each pair is (2m-1)π - 2mπ = -π.So, each pair contributes -π, and there are 1006 pairs, so total from pairs is 1006*(-π) = -1006π.Then, the leftover term is 2013π.So, total sum S = 0 + (-1006π) + 2013π = (2013 - 1006)π = 1007π.Wait, that seems straightforward. Let me verify.Alternatively, think of the sum as:Sum = 0 + (π - 2π) + (3π - 4π) + ... + (2011π - 2012π) + 2013π.Number of pairs: from 1π to 2012π, stepping by 2, so 2012/2 = 1006 pairs, each contributing -π, so total from pairs: -1006π.Then, the last term is 2013π.So, total Sum = -1006π + 2013π = (2013 - 1006)π = 1007π.Yes, that seems correct.Therefore, the sum of all extreme values is 1007π.Wait, just to make sure, let me test with a smaller interval, say [0, 3π].Compute f(0) + f(π) + f(2π) + f(3π).f(0) = 0.f(π) = π.f(2π) = -2π.f(3π) = 3π.Sum = 0 + π - 2π + 3π = (π - 2π) + 3π = (-π) + 3π = 2π.According to the formula, n = 3, so (3 +1)/2 = 2, but wait, 3 is odd, so number of pairs is (3-1)/2 =1 pair, contributing -π, plus the last term 3π, so total sum is -π + 3π = 2π. Which matches.Similarly, for [0, 5π], sum should be 3π.Compute f(0)=0, f(π)=π, f(2π)=-2π, f(3π)=3π, f(4π)=-4π, f(5π)=5π.Sum = 0 + π -2π +3π -4π +5π = (π -2π) + (3π -4π) +5π = (-π) + (-π) +5π = 3π.Which is (5 +1)/2 =3, but wait, 5 is odd, so number of pairs is (5-1)/2=2 pairs, each contributing -π, so total from pairs: -2π, plus last term 5π, total sum: -2π +5π=3π.Yes, that works.So, in general, for [0, nπ], where n is odd, the sum is ((n +1)/2)π? Wait, no, in the case of n=3, sum was 2π, which is (3+1)/2=2, but 2π. Wait, actually, it's (n +1)/2 * π, but in the case of n=3, it's 2π, which is (3+1)/2=2, so 2π. For n=5, it's 3π, which is (5+1)/2=3, so 3π. So, in general, for n odd, sum is ((n +1)/2)π.But in our case, n=2013, which is odd, so sum is (2013 +1)/2 π = 2014/2 π = 1007π.Yes, that's consistent.Therefore, the sum is 1007π.I think that's solid. So, summarizing:(I) The function f(x) is increasing on intervals (2kπ, (2k+1)π) and decreasing on ((2k+1)π, 2(k+1)π) for k = 0,1,2,...(II) The sum of all extreme values on [0, 2013π] is 1007π.**Final Answer**(I) The function is increasing on intervals boxed{(2kpi, (2k+1)pi)} and decreasing on intervals boxed{((2k+1)pi, 2(k+1)pi)} for all non-negative integers k.(II) The sum of all extreme values is boxed{1007pi}.

💡First, I recognize that the problem is asking for the minimum number of people in the urban village of Owlna who have all four luxuries: refrigerator, television, computer, and air-conditioner.To simplify the calculations, I'll assume there are 100 people in Owlna. This assumption doesn't affect the percentages and makes the math easier.Next, I'll list the percentages of people who have each luxury:- 70% have a refrigerator.- 75% have a television.- 90% have a computer.- 85% have an air-conditioner.Since we're looking for the minimum number of people who have all four items, I'll consider the maximum overlap between these groups. This means assuming that the people who have refrigerators, televisions, and air-conditioners are all part of the largest group, which is the 90% who have computers.Finally, to find the minimum number of people who have all four luxuries, I'll look at the smallest percentage among the remaining three items (refrigerator, television, air-conditioner), which is 70%. Therefore, at least 70 people in Owlna have all four items.

💡Okay, so I have this function f(x) defined piecewise: it's x if x is in set P, and -x if x is in set M. P and M are both non-empty subsets of the real numbers. I need to figure out which of the four statements about this function are correct.Let me start by understanding what f(P) and f(M) mean. f(P) is the set of all f(x) where x is in P, so that would just be P itself since f(x) = x for x in P. Similarly, f(M) is the set of all -x where x is in M, so f(M) is like reflecting M over the origin.Now, let's look at each statement one by one.**Statement 1:** If P ∩ M = ∅, then f(P) ∩ f(M) = ∅.Hmm, so if P and M don't overlap, does that mean their images under f don't overlap either? Let me think of an example. Suppose P is all positive real numbers and M is all negative real numbers. Then P ∩ M is empty because they don't overlap. Now, f(P) would be all positive real numbers, and f(M) would be all positive real numbers as well because f(M) = -M, which turns negatives into positives. So f(P) ∩ f(M) would be all positive real numbers, which is definitely not empty. So Statement 1 is false.**Statement 2:** If P ∩ M ≠ ∅, then f(P) ∩ f(M) ≠ ∅.Okay, so if P and M do overlap, does that mean their images under f also overlap? Let's see. If there's an element x in both P and M, then f(x) would be x (since x is in P) and also -x (since x is in M). So for this x, x must equal -x, which implies x = 0. So the only possible overlap is at 0. Therefore, f(P) ∩ f(M) would contain 0, so it's not empty. Therefore, Statement 2 is true.**Statement 3:** If P ∪ M = ℝ, then f(P) ∪ f(M) = ℝ.Alright, if every real number is in either P or M, does that mean every real number is in either f(P) or f(M)? Let me test this. Suppose P is all non-negative real numbers and M is all negative real numbers. Then P ∪ M is ℝ. But f(P) is all non-negative real numbers, and f(M) is all positive real numbers. So f(P) ∪ f(M) is all non-negative real numbers, which doesn't include the negative numbers. So f(P) ∪ f(M) is not ℝ. Therefore, Statement 3 is false.**Statement 4:** If P ∪ M ≠ ℝ, then f(P) ∪ f(M) ≠ ℝ.So if there's at least one real number not in P or M, does that mean there's at least one real number not in f(P) or f(M)? Let's think. Suppose there's a real number y not in P ∪ M. Then y is neither in P nor in M. So f(P) is P, and f(M) is -M. If y is not in P, then y is not in f(P). If y is not in M, then -y is not in f(M). So y is not in f(P) ∪ f(M). Therefore, f(P) ∪ f(M) is missing y, so it's not ℝ. Hence, Statement 4 is true.So, summarizing:- Statement 1: False- Statement 2: True- Statement 3: False- Statement 4: TrueTherefore, the correct statements are 2 and 4.**Final Answer**The correct statements are boxed{2} and boxed{4}.

💡Alright, so I've got this inequality to prove: [left(a^{5}-a^{2}+3right)left(b^{5}-b^{2}+3right)left(c^{5}-c^{2}+3right) geq (a+b+c)^{3}]where ( a, b, c ) are positive real numbers. Hmm, okay. Let me try to figure out how to approach this.First, I notice that the left side is a product of three similar expressions, each involving a variable raised to the fifth power, minus the square of that variable, plus 3. The right side is the cube of the sum of the variables. So, maybe there's a way to relate each of these expressions on the left to something involving ( a, b, c ) individually, and then use an inequality that works on products or sums.I remember that inequalities like AM-GM or Hölder's might be useful here. Let me think about AM-GM first. AM-GM relates the arithmetic mean and the geometric mean, but I'm not sure how to directly apply it here because the expressions on the left are more complicated.Wait, maybe I can find a lower bound for each ( a^5 - a^2 + 3 ). If I can show that each of these terms is at least something, then multiplying those lower bounds together might give me the desired inequality.Let's consider one term, say ( a^5 - a^2 + 3 ). I need to find a lower bound for this expression. Maybe I can factor it or find a way to express it in terms of ( a^3 ) or something else that relates to the right side.Let me try to manipulate ( a^5 - a^2 + 3 ). Maybe I can write it as ( a^5 + 3 - a^2 ). Hmm, not sure if that helps. What if I factor ( a^2 ) out of the first two terms? That would give ( a^2(a^3 - 1) + 3 ). Still not very helpful.Wait, another thought: perhaps I can compare ( a^5 - a^2 + 3 ) to ( a^3 + 2 ). Let me see if ( a^5 - a^2 + 3 geq a^3 + 2 ). If that's true, then each term on the left is at least ( a^3 + 2 ), and then I can work with that.So, let's check if ( a^5 - a^2 + 3 geq a^3 + 2 ). Subtracting ( a^3 + 2 ) from both sides, we get:[a^5 - a^2 + 3 - a^3 - 2 geq 0 implies a^5 - a^3 - a^2 + 1 geq 0]Okay, so I need to show that ( a^5 - a^3 - a^2 + 1 geq 0 ) for all positive real numbers ( a ).Hmm, let's factor this expression. Maybe I can factor by grouping. Let's group ( a^5 - a^3 ) and ( -a^2 + 1 ):[a^3(a^2 - 1) - (a^2 - 1) = (a^3 - 1)(a^2 - 1)]Wait, that's not quite right. Let me check:( a^3(a^2 - 1) = a^5 - a^3 ), and ( - (a^2 - 1) = -a^2 + 1 ). So, yes, that works. So, the expression factors as:[(a^3 - 1)(a^2 - 1)]Now, since ( a ) is a positive real number, let's analyze the sign of each factor.Case 1: ( a > 1 )- ( a^3 - 1 > 0 )- ( a^2 - 1 > 0 )- So, the product is positive.Case 2: ( a = 1 )- ( a^3 - 1 = 0 )- ( a^2 - 1 = 0 )- So, the product is 0.Case 3: ( 0 < a < 1 )- ( a^3 - 1 < 0 )- ( a^2 - 1 < 0 )- So, the product is positive (negative times negative).Therefore, in all cases, ( (a^3 - 1)(a^2 - 1) geq 0 ). So, ( a^5 - a^3 - a^2 + 1 geq 0 ), which implies ( a^5 - a^2 + 3 geq a^3 + 2 ).Great! So, each term on the left is at least ( a^3 + 2 ), ( b^3 + 2 ), and ( c^3 + 2 ) respectively. Therefore, the product on the left is at least:[(a^3 + 2)(b^3 + 2)(c^3 + 2)]Now, I need to show that this is at least ( (a + b + c)^3 ). Hmm, how can I relate ( (a^3 + 2)(b^3 + 2)(c^3 + 2) ) to ( (a + b + c)^3 )?I recall that Hölder's inequality might be useful here. Hölder's inequality states that for positive real numbers and exponents ( p, q ) such that ( 1/p + 1/q = 1 ), the following holds:[sum_{i=1}^n a_i b_i leq left( sum_{i=1}^n a_i^p right)^{1/p} left( sum_{i=1}^n b_i^q right)^{1/q}]But I'm not sure if that's directly applicable here. Maybe another form of Hölder's inequality or another inequality altogether.Wait, another thought: perhaps I can use the AM-GM inequality on each ( a^3 + 2 ). Let's see.The expression ( a^3 + 2 ) can be written as ( a^3 + 1 + 1 ). By AM-GM, we have:[frac{a^3 + 1 + 1}{3} geq sqrt[3]{a^3 cdot 1 cdot 1} = a]Multiplying both sides by 3:[a^3 + 2 geq 3a]Similarly, ( b^3 + 2 geq 3b ) and ( c^3 + 2 geq 3c ).So, multiplying these inequalities together:[(a^3 + 2)(b^3 + 2)(c^3 + 2) geq (3a)(3b)(3c) = 27abc]But I need to relate this to ( (a + b + c)^3 ). Hmm, so I have:[(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) geq (a^3 + 2)(b^3 + 2)(c^3 + 2) geq 27abc]But ( 27abc ) is not necessarily greater than or equal to ( (a + b + c)^3 ). In fact, by AM-GM, ( (a + b + c)^3 geq 27abc ). So, this approach might not directly give me the desired inequality.Wait, maybe I need a different way to apply Hölder's inequality. Let me think about Hölder's inequality in the form:[(a_1 + a_2 + dots + a_n)^k leq (a_1^p + a_2^p + dots + a_n^p)^{k/p} (1 + 1 + dots + 1)^{(k - k/p)}]But I'm not sure if that's helpful here.Alternatively, perhaps I can consider the product ( (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3) ). This looks similar to the form where Hölder's inequality can be applied.In fact, Hölder's inequality in three variables states that:[(a_1 + a_2 + a_3)(b_1 + b_2 + b_3)(c_1 + c_2 + c_3) geq (a_1^{1/3} b_1^{1/3} c_1^{1/3} + a_2^{1/3} b_2^{1/3} c_2^{1/3} + a_3^{1/3} b_3^{1/3} c_3^{1/3})^3]But I'm not sure if that's directly applicable here.Wait, another approach: maybe I can use the fact that ( (a^3 + 1 + 1) geq 3a ) from AM-GM, as I did before, but then think about how to combine these.Alternatively, perhaps I can use the inequality ( (x + y + z)^3 leq (x^3 + y^3 + z^3)(1 + 1 + 1)^2 ) by Hölder's inequality, but I'm not sure.Wait, let me try to think differently. I have:[(a^3 + 2)(b^3 + 2)(c^3 + 2) geq (a + b + c)^3]I need to show this. Let me expand the left side and see if I can compare it to the right side.Expanding ( (a^3 + 2)(b^3 + 2)(c^3 + 2) ):First, multiply ( (a^3 + 2)(b^3 + 2) ):[a^3 b^3 + 2a^3 + 2b^3 + 4]Now multiply this by ( (c^3 + 2) ):[(a^3 b^3 + 2a^3 + 2b^3 + 4)(c^3 + 2) = a^3 b^3 c^3 + 2a^3 b^3 + 2a^3 c^3 + 4a^3 + 2b^3 c^3 + 4b^3 + 4c^3 + 8]So, the expanded form is:[a^3 b^3 c^3 + 2a^3 b^3 + 2a^3 c^3 + 4a^3 + 2b^3 c^3 + 4b^3 + 4c^3 + 8]Now, the right side is ( (a + b + c)^3 ), which expands to:[a^3 + b^3 + c^3 + 3a^2 b + 3a^2 c + 3ab^2 + 3ac^2 + 3b^2 c + 3bc^2 + 6abc]Hmm, comparing these two expressions, it's not obvious how to show that the left side is greater than or equal to the right side. Maybe expanding isn't the best approach.Perhaps I need to use another inequality or a different strategy. Let me think about the original inequality again.We have:[left(a^{5}-a^{2}+3right)left(b^{5}-b^{2}+3right)left(c^{5}-c^{2}+3right) geq (a+b+c)^{3}]And we've established that each ( a^5 - a^2 + 3 geq a^3 + 2 ). So, the left side is at least ( (a^3 + 2)(b^3 + 2)(c^3 + 2) ).Now, I need to show that ( (a^3 + 2)(b^3 + 2)(c^3 + 2) geq (a + b + c)^3 ).Wait, maybe I can use the AM-GM inequality in a different way. Let's consider the terms ( a^3 + 1 + 1 geq 3a ), as before. So, each ( a^3 + 2 geq 3a ). Therefore, the product ( (a^3 + 2)(b^3 + 2)(c^3 + 2) geq 27abc ).But ( (a + b + c)^3 ) is greater than or equal to ( 27abc ) by AM-GM, since:[frac{a + b + c}{3} geq sqrt[3]{abc} implies (a + b + c)^3 geq 27abc]So, we have:[(a^3 + 2)(b^3 + 2)(c^3 + 2) geq 27abc leq (a + b + c)^3]Wait, that's not helpful because ( 27abc leq (a + b + c)^3 ), but we need ( (a^3 + 2)(b^3 + 2)(c^3 + 2) geq (a + b + c)^3 ). So, this approach shows that ( (a^3 + 2)(b^3 + 2)(c^3 + 2) geq 27abc leq (a + b + c)^3 ), which doesn't directly help.Hmm, maybe I need to find another way to relate ( (a^3 + 2)(b^3 + 2)(c^3 + 2) ) to ( (a + b + c)^3 ).Wait, another idea: perhaps I can use the fact that ( a^3 + 2 geq 3a ) and then apply Hölder's inequality in a different form.Hölder's inequality in three variables states that:[(a_1 + a_2 + a_3)(b_1 + b_2 + b_3)(c_1 + c_2 + c_3) geq (a_1^{1/3} b_1^{1/3} c_1^{1/3} + a_2^{1/3} b_2^{1/3} c_2^{1/3} + a_3^{1/3} b_3^{1/3} c_3^{1/3})^3]But I'm not sure how to apply this here.Wait, maybe I can think of ( (a^3 + 1 + 1) ) as ( a^3 + 1 + 1 ) and similarly for ( b ) and ( c ). Then, using Hölder's inequality on these three terms.Let me try that. Let me set:- For the first term: ( x_1 = a^3 ), ( x_2 = 1 ), ( x_3 = 1 )- Similarly for ( b ) and ( c ).Then, applying Hölder's inequality:[(x_1 + x_2 + x_3)(y_1 + y_2 + y_3)(z_1 + z_2 + z_3) geq (x_1^{1/3} y_1^{1/3} z_1^{1/3} + x_2^{1/3} y_2^{1/3} z_2^{1/3} + x_3^{1/3} y_3^{1/3} z_3^{1/3})^3]But in this case, ( x_1 = a^3 ), ( y_1 = 1 ), ( z_1 = 1 ), and similarly for the other terms. Wait, no, that might not be the right way to set it up.Alternatively, maybe I can consider each ( a^3 + 1 + 1 ) as a sum of three terms and apply Hölder's inequality accordingly.Wait, another thought: perhaps I can use the inequality ( (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3) geq (a + b + c)^3 ) by Hölder's inequality.Let me check that. If I set up the terms as follows:- First factor: ( a^3 + 1 + 1 )- Second factor: ( 1 + b^3 + 1 )- Third factor: ( 1 + 1 + c^3 )Then, applying Hölder's inequality with exponents ( p = q = r = 3 ), we get:[(a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3) geq (a + b + c)^3]Yes, that seems right! Because Hölder's inequality in this form states that:[left( sum_{i=1}^n x_i right) left( sum_{i=1}^n y_i right) left( sum_{i=1}^n z_i right) geq left( sum_{i=1}^n (x_i y_i z_i)^{1/3} right)^3]In our case, each sum has three terms, and the product inside the cube root would be ( a^3 cdot 1 cdot 1 ), ( 1 cdot b^3 cdot 1 ), and ( 1 cdot 1 cdot c^3 ), which simplifies to ( a, b, c ). Therefore, the inequality becomes:[(a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3) geq (a + b + c)^3]Which is exactly what we needed.So, putting it all together:1. We showed that ( a^5 - a^2 + 3 geq a^3 + 2 ) for all positive real ( a ).2. Similarly, ( b^5 - b^2 + 3 geq b^3 + 2 ) and ( c^5 - c^2 + 3 geq c^3 + 2 ).3. Therefore, the product on the left is at least ( (a^3 + 2)(b^3 + 2)(c^3 + 2) ).4. We then applied Hölder's inequality to show that ( (a^3 + 2)(b^3 + 2)(c^3 + 2) geq (a + b + c)^3 ).Thus, combining these results, we have:[left(a^{5}-a^{2}+3right)left(b^{5}-b^{2}+3right)left(c^{5}-c^{2}+3right) geq (a + b + c)^3]Which is the inequality we wanted to prove. So, I think this approach works!