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

∑k=1n1xk+1=1.\sum_{k=1}^n\frac{1}{x_k+1}=1.

Show that

∑k=1n1xk≤1n−1(∑k=1nxk).\sum_{k=1}^n\frac{1}{\sqrt{x_k}}\leq\frac{1}{n-1}\left(\sum_{k=1}^n\sqrt{x_k}\right).

#Algebra

Note by Victor Loh
6 years, 9 months ago

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Comments

Solution: It suffices to show that

n(∑k=1n1xk)≤∑k=1n(xk+1xk),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(∑k=1n1xk)≤(∑k=1nxk+1xk)(∑k=1n1xk+1).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)=x+1x=x+1x,x∈(0,+∞).f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{x+1}{\sqrt{x}}, x\in(0,+ \infty). It is easy to verify that ff is non-decreasing on (1,+∞)(1,+\infty) and that f(x)=f(1x)f(x)=f\left(\frac{1}{x}\right) for all x>0.x>0. Furthermore, from the given conditions, it follows that only x1x_1 can be less than 11 and that 1x2+1≤1−1x1+1=x1x1+1.\frac{1}{x_2+1} \leq 1-\frac{1}{x_1+1}=\frac{x_1}{x_1+1}. Hence, 1x1≤x2.\frac{1}{x_1}\leq x_2. It is now clear that (for both cases 1≤x11\leq x_1 and x<1x<1) that

f(x1)=f(1x1)≤f(x2)≤…≤f(xn).f(x_1)=f\left(\frac1{x_1}\right) \leq f(x_2)\leq \ldots \leq f(x_n).

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

n(∑k=1n1xk)≤(∑k=1nxk+1xk)(∑k=1n1xk+1),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 1x1+1=…=1xn+1,\frac{1}{x_1+1} = \ldots = \frac{1}{x_n+1}, or x1+1x1=…=xn+1xn,\frac{x_1+1}{\sqrt{x_1}}=\ldots = \frac{x_n+1}{\sqrt{x_n}}, which implies that x1=x2=…=xn.x_1=x_2=\ldots=x_n. Thus equality holds if and only if x1=x2=…=xn=n−1.□x_1=x_2=\ldots=x_n=n-1._\square

Victor Loh - 6 years, 9 months ago
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