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Question: Let $x_1, x_2, ... , x_m$ be non negative numbers, and let $\sum_{i=1}^{m}x_i=k.$ If $s>1,$ then

$$\sum_{i=1}^{m}x_i^s \geq \frac{k^s}{m^{s-1}}.$$ Equality holds iff $x_i=k/m, i=1 , 2 , ... , m.$

I am trying to prove the given inequality using induction or A.M and G.M inequalities but am not able to prove it.

Any hint or solution is appreciated Thank you!

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    $\begingroup$ Just apply Jensen's inequality to the convex function $x \mapsto x^s$. $\endgroup$
    – Martin R
    Commented May 4, 2022 at 7:06
  • $\begingroup$ @MartinR can it prove using the convex function $x^s$ definition and property? $\endgroup$ Commented May 12, 2022 at 8:40
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    $\begingroup$ @ManishSaini This is just the power means inequality $\;\displaystyle \left(\frac{\sum_{i=1}^{m}x_i^s}{m}\right)^{1/s} \geq \frac{\sum_{i=1}^{m}x_i}{m}\,$. $\endgroup$
    – dxiv
    Commented May 14, 2022 at 3:31

2 Answers 2

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Jensen's inequality: If $\lambda_1,$ $\lambda_2,$ $...,$ $\lambda_n$ be non negative real numbers such that $\lambda_1+\lambda_2+...+\lambda_n=1,$ and $f$ is convex function then $f(\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n) \leq \lambda_1f(x_1)+\lambda_nf(x_n)+...+\lambda_nf(x_n),$ for any $x_1,x_2,...,x_n.$

Let $\phi :$ $x \to x^s$ be a convex function for $x>0$ and $s>1$ then,

By Jensen's inequality

$\phi((1/k)x_1+(1/k)x_2+...+(1/k)x_n) \leq (1/k) \phi(x_1)+(1/k) \phi(x_2)+...+(1/k) \phi(x_n)$ $\implies ((1/k)x_1+(1/k)x_2+...+(1/k)x_n)^s \leq (1/k)x_1^s+(1/k)x_2^s+...(1/k)x_n^s$ $\implies (1/k)^s(x_1+x_2+...+x_n)^s \leq (1/k)(x_1^s+x_2^s+...+x_n^s)$ $\implies (x_1^s+x_2^s+...+x_n^s) \geq \frac{(x_1+x_2+...+x_n)^s}{k^{s-1}}.$

This is my approach to proving the given inequality using the convex function and Jensen's inequality.

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We will try proving $x_1^s+x_2^s \geq 2(\frac {x_1+x_2} 2)^s$ with equality happens iff $x_1=x_2$. We see that the inequality is homogeneous so we let $x_1+x_2=2$ and substitute $x_1=1+l$ and $x_2=1-l$ then we have $(1+l)^s+(1-l)^s \geq 2$ with equality happens only when $l=0$ or $x_1=x_2$ so it is true. Say we have a list of values of $x_i$ which we order from smallest to largest. If all of the values are not equal, we can replace the value of the smallest and largest $x_i$, each with their arithmetic mean then the sum we get will be smaller than the previous sum such that the boundary of the smallest and highest value shrink and still satisfying the sum condition of the problem. If we keep applying this, the difference between the smallest and highest value will approach 0, this means the minimum should happen when all $x_i$ are infinitely close to each other and $x_i^s$ is continuous, so we should have minimum only when all $x_i$ are equal which gives the inequality.

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  • $\begingroup$ basically smoothing inequality $\endgroup$
    – Hi there
    Commented May 3, 2022 at 18:51
  • $\begingroup$ @Hithere Do you have some references/links/books related to Smoothing inequality then please share with me. $\endgroup$ Commented May 4, 2022 at 5:37
  • $\begingroup$ @ManishSaini yisun.io/notes/inequalities.pdf $\endgroup$
    – Hi there
    Commented May 4, 2022 at 8:45

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