### Proving inequalities using convex functions

I have found that I am pretty bad at finding “clever factors” for Cauchy-Schwarz, whose bounds can be known from the given conditions. However, I am slowly getting comfortable with the idea of converting the expression into a convex function, and then using the Majorization Theorem.

(Turkey) Let $n\geq 2$, and $x_1,x_2,\dots,x_n$ positive reals such that $x_1^2+x_2^2+\dots+x_n^2=1$. Find the minimum value of $\sum\limits_{i=1}^n \frac{x_i^5}{\sum\limits_{j\neq i} x_j}$

My proof: We have $(x_1^2+1)+\dots+(x_n^2+1)=n+1$. Hence, using AM-GM inequality, we have $x_1+\dots+x_n\leq \frac{n+1}{2}$.

The expression we finally get is

$\sum\limits_{i=1}^n \frac{x_i^5}{\sum\limits_{j\neq i} x_j}\geq \sum\limits_{i=1}^n \frac{x_i^5}{\frac{n+1}{2}-x_i}$

Consider the function $f(x)=\frac{x^{5/2}}{\frac{n+1}{2}-\sqrt{x}}$. By differentiating twice, we know that for $x\leq 1$, this function is convex.

Hence, $f(x_1^2)+\dots+f(x_n^2)$ will be minimized only when $x_1^2=\dots=x_n^2$, which we know from the conditions given in the question, is $\frac{1}{n}$.

Note that

$f(x_1^2)+\dots+f(x_n^2)= \sum\limits_{i=1}^n \frac{x_i^5}{\frac{n+1}{2}-x_i}$

Hence, the minimum is attained when $x_1=\dots=x_n=\frac{1}{\sqrt{n}}$, and is equal to $\frac{1}{n(n-1)}$, which is found by substituting $x_1=\frac{1}{\sqrt{n}}$ in the original expression.