Not As Simple As AM-GM.

Problem: Let \(0<x_1\leq x_2\ldots\leq x_{n-2}\leq x_{n-1}\leq x_n\) with \(n\geq 2\) and

\[\sum_{k=1}^n\frac{1}{x_k+1}=1.\]

Show that

\[\sum_{k=1}^n\frac{1}{\sqrt{x_k}}\leq\frac{1}{n-1}\left(\sum_{k=1}^n\sqrt{x_k}\right).\]

Note by Victor Loh
3 years, 8 months ago

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Solution: It suffices to show that

\[n\left(\sum_{k=1}^n\frac{1}{\sqrt{x_k}}\right)\leq\sum_{k=1}^n\left(\sqrt{x_k}+\frac{1}{\sqrt{x_k}}\right),\]

or equivalently

\[n\left(\sum_{k=1}^n\frac{1}{\sqrt{x_k}}\right)\leq\left(\sum_{k=1}^n\frac{x_k+1}{\sqrt{x_k}}\right)\left(\sum_{k=1}^n\frac{1}{x_k+1}\right).\]

Consider the function \(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{x+1}{\sqrt{x}}, x\in(0,+ \infty).\) It is easy to verify that \(f\) is non-decreasing on \((1,+\infty)\) and that \(f(x)=f\left(\frac{1}{x}\right)\) for all \(x>0.\) Furthermore, from the given conditions, it follows that only \(x_1\) can be less than \(1\) and that \(\frac{1}{x_2+1} \leq 1-\frac{1}{x_1+1}=\frac{x_1}{x_1+1}.\) Hence, \(\frac{1}{x_1}\leq x_2.\) It is now clear that (for both cases \(1\leq x_1\) and \(x<1\)) that

\[f(x_1)=f\left(\frac1{x_1}\right) \leq f(x_2)\leq \ldots \leq f(x_n).\]

This means that the sequence \(\left(\frac{x_k+1}{x_k}\right)_{k=1}^n\) is non-decreasing. Thus, by Chebyshev's inequality, we have

\[n\left(\sum_{k=1}^n\frac{1}{\sqrt{x_k}}\right)\leq\left(\sum_{k=1}^n\frac{x_k+1}{\sqrt{x_k}}\right)\left(\sum_{k=1}^n\frac{1}{x_k+1}\right),\]

and we are done.

Equality is attained if and only if \(\frac{1}{x_1+1} = \ldots = \frac{1}{x_n+1},\) or \(\frac{x_1+1}{\sqrt{x_1}}=\ldots = \frac{x_n+1}{\sqrt{x_n}},\) which implies that \(x_1=x_2=\ldots=x_n.\) Thus equality holds if and only if \(x_1=x_2=\ldots=x_n=n-1._\square\)

Victor Loh - 3 years, 8 months ago

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