Proving inequality using Lagrange multipliers

Question

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

Answer 1

Unless you restrict the problem to $n \geq 1$ the statement is false.

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

Answer 2

We assume that $\sum_{i=0}^{n}x_i=n$. Thus, we would have to show that $\prod_{i=0}^{n}x_i \leq 1$. Consider the set $[0,n]^n$. Since, it is closed and bounded, by Heine-Borel Theorem it is compact. Let $f(x_1,x_2, \cdots, x_n)=\prod_{i=0}^{n}x_i$ (what we want to optimize) and $g(x_1, x_2, \cdots , x_n)=\sum_{i=0}^{n}x_i$ (our constraint) be two continuous functions. Now we find the gradient vectors of $f$ and $g$. Taking the partial derivative of $f$ and $g$ given that $\nabla g$ can never be zero, we get, $$\nabla f=\big \langle (x_2\cdots x_n), (x_1x_3\cdots x_n), \cdots , (x_1\cdots x_{n-1}) \big \rangle$$ and, $$ \nabla g= \big \langle 1,1,\cdots ,1 \big \rangle$$ which are also continuous. Now from Lagrange Multiplier we get a system of $n$ equations, $$x_2x_3\cdots x_n=\lambda \\ x_1x_3\cdots x_n=\lambda \\ \vdots \\ x_1x_2\cdots x_{n-1}=\lambda$$ \end{center} Thus, $\frac{x_j}{x_k}=1$ $\forall$ $j,k \in 1,2,\cdots n$ whcih implies that $x_i=1$ where it's fairly easy to see that AM-GM holds. Now all that remains to check the edge cases. \ When some $x_m=n$, then due to our constraint all other $x_i$ have to be $0$. AM-GM clearly holds in this case since $1>0$. Similarly if some $x_n=0$, then our RHS is $0$ and LHS greater than $1$, which again clearly works.