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Let $x_{1},x_{2},\cdots,x_{n}$ be postive real numbers, and such $x_{1}x_{2}\cdots x_{n}=1$, Show that $$\dfrac{1}{x_{1}(x_{1}+1)}+\dfrac{1}{x_{2}(x_{2}+1)}+\cdots+\dfrac{1}{x_{n}(x_{n}+1)}\ge\dfrac{n}{2}$$

I try use Cauchy-Schwarz inequality.but I don't see how to make the right estimates.If anyone has an idea how to proceed in that inequality

for $n=3$,even it is hard to prove it

$x_{1}=\dfrac{a}{b},x_{2}=\dfrac{b}{c},x_{3}=\dfrac{c}{a}$ It suceffent to prove $$\sum_{cyc}\dfrac{b^2}{a(a+b)}\ge\dfrac{3}{2}$$ Use Cauchy-Schwarz inequality we have $$\sum_{cyc}\dfrac{b^2}{a(a+b)}\sum_{cyc}a(a+b)\ge (a+b+c)^2$$ $$\Longleftrightarrow 2(a+b+c)^2\ge 3\sum_{cyc}(a^2+ab)$$ this inequality is wrong

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Let $x_i = e^{y_i}$. We want to find: $$ \inf_{\sum y_i=0}\sum_{i}\frac{1}{e^{y_i}+e^{2y_i}} $$ but since $f(x)=\frac{1}{e^{x}+e^{2x}}$ is a convex function$\phantom{}^{(*)}$ on $\mathbb{R}$, $$ \sum_{i} f(y_i) \geq n\cdot f\left(\frac{1}{n}\sum_{i} y_i\right)= n\cdot f(0)=\frac{n}{2} $$ by Jensen's inequality, with equality attained at $y_1=y_2=\ldots=0$, and the previous $\inf$ is indeed a minimum.

Proof of $(*)$. $$\frac{d^2}{dx^2}\,f(x) = \frac{3+3e^x+2\cosh(x)}{(e^x+1)^3} \geq \frac{1}{(1+e^x)^2} > 0.$$

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  • $\begingroup$ Nice !Thanks I have known it $\endgroup$
    – inequality
    May 31 '16 at 15:04
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Let $y_i=\log x_i$. Then, $\sum_{i=1}^ny_i=0$. Define $$f(x)=\frac1{e^x(e^x+1)}$$ $$f'(x)=-{e^{-x} (1+2 e^x)\over(1+e^x)^2}$$ $$f''(x)={e^{-x} (1+3 e^x+4 e^{2 x})\over(1+e^x)^3}>0$$ So, $f$ is convex. Thus, by Jensen's Inequality $$\sum_{i=1}^nf(y_i)\ge nf(\sum_{i=1}^ny_i)=n\cdot f(0)$$ $$\implies \sum_{i=1}^n\frac1{x_i(x_i+1)}\ge \frac{n}2$$

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Let $x_i=\frac{a_i}{a_{i+1}}$, where $a_i>0$ and $a_{n+1}=a_{1}$. We have $$\tag{1}\sum_{cyc}\frac{1}{x_i(x_i+1)}=\sum_{cyc}\frac{1}{\frac{a_i}{a_{i+1}}\cdot\left( \frac{a_i}{a_{i+1}}+1\right)}=\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot\left(a_i+a_{i+1}\right)}$$

Notice that: $$\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot (a_i+a_{i+1})}-\sum_{cyc}\frac{a_{i}^2}{a_i\cdot (a_i+a_{i+1})}=\sum_{cyc}\frac{(a_{i+1}-a_i)\cdot(a_{i+1}+a_i)}{a_i\cdot(a_i+a_{i+1})}=\sum_{cyc}\left(\frac{a_{i+1}}{a_i}-1\right)=\sum_{cyc}\frac{a_{i+1}}{a_i}-n\geqslant n\sqrt[n]{\prod_{cyc}\frac{a_{i+1}}{a_i}}-n\geqslant 0$$ Thus $$\tag{2}\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot (a_i+a_{i+1})} \geqslant\sum_{cyc}\frac{a_{i}^2}{a_i\cdot (a_i+a_{i+1})}$$

From $(1)$ and $(2)$, we have $$\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot(a_i+a_{i+1})}\geqslant \frac12\sum_{cyc}\frac{a_i^2+a_{i+1}^2}{a_i\cdot(a_i+a_{i+1})}\geqslant\frac14\sum_{cyc}\frac{(a_i+a_{i+1})^2}{a_i\cdot(a_i+a_{i+1})}=\frac14\sum_{cyc}\frac{a_i+a_{i+1}}{a_i}=\frac14\sum_{cyc}\left(1+\frac{a_{i+1}}{a_i}\right)=\frac14\left(n+\sum_{cyc}\frac{a_{i+1}}{a_i}\right)\geqslant \frac14\cdot(2n)=\frac{n}{2}$$ QED.

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