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Prove that if $x_i > 0$ for all $i$ then

$$\begin{align*} &(x_1^{19}+x_2^{19}+\cdots+x_n^{19})(x_1^{93}+x_2^{93}+\cdots+x_n^{93})\\ &\geq (x_1^{20}+x_2^{20}+\cdots+x_n^{20})(x_1^{92}+x_2^{92}+\cdots+x_n^{92}). \end{align*}$$

and find when equality holds.

I'm not sure how to start...I tried maybe using induction, but I wasn't sure how. I also thought about using the Mean Inequality Chain and Cauchy, but I couldn't figure out how to use it. Any help on how to get started would be greatly appreciated, thanks!

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First we prove the inequality for $2$ positive real numbers $x,y > 0$, and for $m,n \geq 2$, and $n-1 > m$:

$(x^m+y^m)(x^n+y^n) \geq (x^{m+1}+y^{m+1})(x^{n-1}+y^{n-1}) \iff (y-x)(x^my^{n-1}-x^{n-1}y^m) \geq 0$. (*)

Observe that $y > x \iff \dfrac{y}{x} > 1 \iff \left(\dfrac{y}{x}\right)^{n-1-m} > 1 \iff x^my^{n-1}-x^{n-1}y^m > 0$. Thus (*) is true.

Thus using $m = 19, n = 93$, we can expand both sides and regroup the terms into groups of differences of products of the form given in (*), and we're done.

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  • $\begingroup$ Could you elaborate on how you expand and regroup? I just can't figure it out! $\endgroup$ – bob the pie Aug 20 '15 at 6:15
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Let $f(k) = \log(x_1^k+x_2^k+\dots+x_n^k)$. Now $x_i^k$ is a convex function of $k$, sum of convex functions is convex, and as $\log$ is increasing, $f(k)$ must also be convex. Hence using Karamata's inequality on $f$, we get $$(a, b) \succ (c, d) \iff f(a) + f(b) \ge f(c) + (d)$$ In our inequality, we note $(19, 93) \succ (20, 92)$ and the rest follows.

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Your inequality follows immediately from Muirhead because $(93,19,0)\succ(92,20,0)$.

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