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.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics.

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