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I have this question. Prove that for all $ x,y\geq 0 $, $$ \dfrac{x^n+y^n}{2}\geq \bigg(\dfrac{x+y}{2}\bigg)^n $$ using the method of Lagrange Multipliers, via $$ \min \dfrac{x^n+y^n}{2}, \text{where $x+y=s$} $$ for some $s\geq 0$.

This is what I did. I consider $f(x,y)=\dfrac{x^n+y^n}{2}$ and the constraint $x+y=s$.

Then, I get $\begin{cases} \dfrac{nx^{n-1}}{2}=\lambda\\ \dfrac{ny^{n-1}}{2}=\lambda \end{cases}$. Using that system, I get that $x=y$. Going back to the constraint, I get $$2x=2y=s\ i.e.\ x=y=\dfrac{s}{2}.$$ I think I almost get the answer since $\bigg(\dfrac{s}{2}\bigg)^n=\bigg(\dfrac{x+y}{2}\bigg)^n$ but I get lost.

Need help. Thank you so much

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    $\begingroup$ Remember that you are seeking to minimize $ \ \dfrac{x^n+y^n}{2} \ $ : is that greater than or equal to $ \ \frac{s^n}{2^n} \ $ when you insert the values of $ \ x \ $ and $ \ y \ $ that you obtained? (It is.) $\endgroup$ – colormegone Jul 8 '16 at 19:22
  • $\begingroup$ I get $\dfrac{x^n+y^n}{2}\geq \dfrac{s}{2}$ not $\bigg(\dfrac{s}{2} \bigg)^n$. That's why I get lost :( $\endgroup$ – Jax Jul 8 '16 at 19:24
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    $\begingroup$ The function value is $$ \dfrac{\left(\frac{s}{2}\right)^n+\left(\frac{s}{2}\right)^n}{2} \ \ . $$ What does that simplify to? The right-hand side of the (improper) inequality is $$ \left( \dfrac{\left(\frac{s}{2}\right) +\left(\frac{s}{2}\right) }{2} \right)^n \ \ . $$ $\endgroup$ – colormegone Jul 8 '16 at 19:25
  • $\begingroup$ ahhhhh, thank you so much. I got it. I did it in a wrong way. We have $f(x,y)\geq f(\dfrac{s}{2},\dfrac{s}{2})$. I was thinking that we get $f(x,y)\geq \dfrac{s}{2}$. Thank you so much. $\endgroup$ – Jax Jul 8 '16 at 19:30
  • $\begingroup$ I would suggest that you show that x = y actually minimizes your objective function and does not actually maximize it. $\endgroup$ – Doug M Jul 8 '16 at 19:41
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Unless you restrict the problem to $n \geq 1$ the statement is false.

Consider $x=1, y=2, n=\frac12$: $$ \sqrt{2} < \frac32 \implies 1+2+2\sqrt{2} < 6 \implies (1+\sqrt{2})^2 < \sqrt{6}^2\\ \implies 1+\sqrt{2}<\sqrt{6} \implies \frac{1+\sqrt{2}}{2} < \sqrt{\frac{3}{2}} \implies \frac{1^n+2^n}{2} < \left( \frac{1+2}{2}\right)^n $$

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  • $\begingroup$ I think that $n$ is a positive integer. $\endgroup$ – InsideOut Jul 8 '16 at 19:36
  • $\begingroup$ You are right. In my case, $n$ is an integer $\geq 0$. $\endgroup$ – Jax Jul 8 '16 at 19:37
  • $\begingroup$ In addition to that $n$ has to be less than $0$ as well. $\endgroup$ – StubbornAtom Jul 8 '16 at 20:04

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