# Proving inequality with non-negative reals

Let $a_1 , a_2 , ... , a_n$ be $n$ non-negative real numbers all less than $1$ and satisfying

let $$a=\sqrt {\frac1{n}\sum_{i=1}^na^2_i}≥\dfrac{\sqrt{3}}{3}$$ then how do we prove that $$\frac{a_1}{1-a^2_1} + \frac{a_2}{1-a_2^2}+....+\frac{a_n}{1-a_n^2}≥\frac{na}{1-a^2}$$

• This inequality can use RCF-Theorem solve it. – math110 Oct 21 '13 at 8:28
• I think first is $\dfrac{a_{1}}{1-a^2_{1}}$ – math110 Oct 21 '13 at 8:40
• – math110 Oct 21 '13 at 8:47

let $a^2_{i}=x_{i},1\le i\le n$,then

if $x_{1},x_{2},\cdots,x_{n}<1$, are non-negative real numbers,such $$\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}=r\ge\dfrac{1}{3}$$ then we have $$\dfrac{\sqrt{x_{1}}}{1-x_{1}}+\dfrac{\sqrt{x_{2}}}{1-x_{2}}+\cdots+\dfrac{\sqrt{x_{n}}}{1-x_{n}}\ge\dfrac{n\sqrt{r}}{1-r}$$

Proof

Apply RCF-Theorem to the function $$f(u)=\dfrac{\sqrt{u}}{1-u},0\le u\le 1$$ then $$f'(u)=\dfrac{3u^2+6u-1}{4u\sqrt{u}(1-u)^2}$$ it follows that $f$ is convex on $[\dfrac{2}{\sqrt{3}}-1,1)$,since $s=\dfrac{1}{3}>\dfrac{2}{\sqrt{3}}-1$,the function $f$ is convex on $[s,1)$

By RCF-Theorem,

it suffices to show that $g(x)\le g(y)$ for $0\le x\le s\le y<1$ and $x+(n-1)y=ns$

where $$g(t)=\dfrac{f(t)-f(s)}{t-s}$$ for convenience,let $$a=\sqrt{x},b=\sqrt{y},c=\sqrt{s}$$ we have $$g(t^2)=\dfrac{f(t^2)-f(c^2)}{t^2-c^2}=\dfrac{1+ct}{(1-c^2)(1-t^2)(t+c)}$$ and $$g(x)-g(y)=g(a^2)-g(b^2)=(a^2-b^2)\dfrac{a^2+b^2+c(a+b)+c^2-1+ab(1+c^2)+abc(a+b)}{(1-c^2)(1-a^2)(1-b^2)(a+c)(b+c)}$$

since \begin{align*} &a^2+b^2+c(a+b)+c^2-1+ab(1+c^2)+abc(a+b)\\ &\ge a^2+b^2+c(a+b)+c^2-1\\ &\ge a^2+b^2+c\sqrt{a^2+b^2}+c^2-1 \end{align*} it is enought to show that $$x+y+\sqrt{s(x+y)}+s-1\ge 0$$ Indeed,we have $$x+y=\dfrac{ns+(n-2)s}{n-1}\ge\dfrac{ns}{n-1}$$ and therefore $$x+y+\sqrt{s(x+y)}+s-1\ge\left(\dfrac{n}{n-1}+\sqrt{\dfrac{n}{n-1}+1}\right)s-1\ge 3s-1=0$$ Equality occurs only for $x_{1}=x_{2}=\cdots=x_{n}=r$