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Let positive real numbers $p$, $q$, $m$ and $n$ satisfy $p+q=1$ and $m+n=1$, and for real $\alpha$ when

  1. $\alpha\in[-1,0]$ \begin{align} p(np)^\alpha+q(mq)^\alpha\le (mn)^\alpha \end{align}
  2. $\alpha\in(-\infty,-1)\cup(0,\infty)$ \begin{align} (mn)^\alpha\le p(np)^\alpha+q(mq)^\alpha \end{align}

I proved them by applying Bernoulli's inequalities, but what I am looking for is something simpler for lower graders to understand. The expansion (substituting $q=1-p$ and $m=1-n$) by binomials has been tried but the factorial form cannot be explicitly utilized for me.

MathArt
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  • Is Jensen inequality allowed? – Nithuya Mar 11 '22 at 00:08
  • https://yourimageshare.com/ib/2Gruh3alEG Here is how I would do it with Jensen, it only uses the simple inequality with 2 coefficients (p and 1-p) It's maybe still too complicated, though Jensen is just seeing that the segment between two points of the curve is always higher than the curve in a convex function, and then just translate this fact algebraically for each point of the interior of the segment – Nithuya Mar 11 '22 at 00:21
  • @Nithuya, it is will be rude to exclude Jensen's inequality, and your approach is one nice answer which I myself haven't figured it out before. Sorry! – MathArt Mar 11 '22 at 08:00
  • I'm happy to help! – Nithuya Mar 11 '22 at 16:48

1 Answers1

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Rewriting first inequality, dividing it by $(mn)^{\alpha}$ and then changing the exponents becomes: $$p(\frac{m}{p})^{-\alpha} + q(\frac{n}{q})^{-\alpha} \leq 1$$ Note that if $-1 \leq \alpha \leq 0$ then $0 \leq -\alpha \leq 1$ and function $x^{-\alpha}$ is concave, which means using Jensen inequality with $f(x) = x^{\alpha}$ and coefficients $p,q$ (we have $p+q = 1$): $$p(\frac{m}{p})^{-\alpha} + q(\frac{n}{q})^{-\alpha} \leq (p\frac{m}{p} + q\frac{n}{q})^{-\alpha} = (m+n)^{-\alpha} = 1$$

The other inequality is the exact same reasonning but knowing $x^{-\alpha}$ is convex if $-\alpha \not \in [0,1]$.

Nithuya
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