# 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
4 years 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 $$x_1=x_2=\ldots=x_n=n-1._\square$$

- 4 years ago