prove using Lagrange multipliers that for $x,y>0,\space n\in \mathbb N,\space (\frac{x+y}2)^n \leq \frac{x^n+y^n}2 $

Question

I have been asked to prove using Lagrange multipliers that for

\begin{equation*} \space (\frac{x+y}2)^n \leq \frac{x^n+y^n}2,~x,y>0,~n\in \mathbb {N} \end{equation*} I am familiar with the proof by induction, but I can't think of a way to prove it using Lagrange multipliers. I tried working with the function

\begin{equation*} f(x,y)=\frac{x^n+y^n}2-(\frac{x+y}2)^n \end{equation*}

but couldn't figure how to constrain it to $\{(x,y):x,y>0\}$. Maybe the constrain should be $g(x,y)=\sqrt x +\sqrt y -r$ for a given $r>0$ and then prove it for all $r$?

Answer 1

Let $a=\frac{x}{x+y},b=\frac{y}{x+y}$. The inequality $$(\frac{x+y}2)^n \leq \frac{x^n+y^n}2$$ is equivalent to $$\frac{a^n+b^n}2\ge\frac{1}{2^n} $$ for $a,b>0$ and $a+b=1$. Let $$ f(a,b,\lambda)=\frac{a^n+b^n}2-\lambda[(\frac{a+b}2)^n-1]. $$ Then solving $$ \frac{\partial f}{\partial a}=\frac n2a^{n-1}-\frac{\lambda n}{2^n}(a+b)^{n-1}=0,\frac{\partial f}{\partial b}=\frac n2b^{n-1}-\frac{\lambda n}{2^n}(a+b)^{n-1}=0,a+b=1$$ gives $$ a=b=\frac12,\lambda=1. $$ Thus $\frac{a^n+b^n}2$ reaches the minimum $\frac{1}{2^n}$ when $a=b=\frac12$ or $$ \frac{a^n+b^n}2\ge \frac{1}{2^n}. $$ So $$ \frac{x^n+y^n}2\ge (\frac{x+y}{2})^n. $$

Answer 2

Hint : Calculate the maximum of $(\frac{x+y}{2})^n$ and the minimum of $\frac{x^n+y^n}{2}$ with Lagrange multipliers and compare the results.