Proving inequality using Lagrange multipliers 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

Unless you restrict the problem to $n \geq 1$ the statement is false.
Consider $x=1, y=2, n=\frac12$:
$$
\sqrt{2} < \frac32 \implies 1+2+2\sqrt{2} < 6 \implies (1+\sqrt{2})^2 < \sqrt{6}^2\\ \implies 1+\sqrt{2}<\sqrt{6} \implies \frac{1+\sqrt{2}}{2} < \sqrt{\frac{3}{2}} \implies  \frac{1^n+2^n}{2} < \left( \frac{1+2}{2}\right)^n
$$
