let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $ then prove that : 
Let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $. Prove that :
  $$\frac{ a_{1} }{ a_{2} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\frac{ a_{2} }{ a_{1} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\cdots+\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}}
 \geq \frac{ 4}{ a_{1}+ a_{2} + a_{3}+\cdots+ a_{n}}$$


I do not know where to start. please help me .
 A: I think we can start by C-S:
$$\sum_{k=1}^n\frac{a_k}{\sum\limits_{i\neq k}a_i^2}=\sum_{k=1}^n\frac{a_k^2}{a_k\sum\limits_{i\neq k}a_i^2}\geq\frac{\left(\sum\limits_{k=1}a_k\right)^2}{\sum\limits_{k=1}^na_k\sum\limits_{i\neq k}a_i^2}.$$
Thus, it remains to prove that
$$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limits_{k=1}^n\left(a_k\sum\limits_{i\neq k}a_i^2\right)$$ or
$$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limits_{k=1}^n\left(a_k^2\sum\limits_{i\neq k}a_i\right).$$
Now, let $\sum\limits_{k=1}^na_k=1$. 
Hence, we need to prove that $$\sum\limits_{k=1}^nf(a_k)\leq\frac{1}{4},$$
where $f(x)=x^2-x^3$.
But $f''(x)=2-6x$, which says that $f$ has an unique inflection point on $(0,1)$.
Thus, by Vasc's HCF Theorem it's enough to prove our inequality for
$a_1=a_2=...=a_{n-1}=x$ and $a_n=1-(n-1)x$, where $0<x<\frac{1}{n-1}$.
Thus, it's enough to prove that $g(x)\geq0,$ where
$$g(x)=\frac{1}{4}-(n-1)(x^2-x^3)-(1-(n-1)x)^2+(1-(n-1)x)^3.$$
We have
$$g'(x)=(n-1)(1-nx)(3(n-2)x-1).$$
We see that $0<\frac{1}{3(n-2)}<\frac{1}{n}<\frac{1}{n-1}$, $x_{min}=\frac{1}{3(n-2)}$ and $x_{max}=\frac{1}{n}$, which says
$$g(x)\geq\min\left\{g\left(\frac{1}{3(n-2)}\right),g\left(\frac{1}{n-1}\right)\right\}=\min\left\{\frac{11n^2-56n+72}{108(n-2)^2},\frac{(n-3)^2}{4(n-1)^2}\right\}\geq0.$$
Done!
For the proof we can use also the Vasc's EV Method.
A: Notice that $$\left( \frac{a_1}{a_2^2 + a_3^2 + ... + a_n^2} + \frac{a_2}{a_1^2 + a_3^2 + ... + a_n^2} + \ ... \ \frac{a_n}{a_1^2 + a_2^2+... + a_{n-1}^2}\right)(a_1 + a_2 + ... + a_n) \ge \left( \sqrt{\frac{a_1^2}{a_2^2 + a_3^2 + ... + a_n^2}} + \sqrt{\frac{a_2^2}{a_1^2 + a_3^2 + ... + a_n^2}} + \ ... \ \sqrt{\frac{a_n^2}{a_1^2 + a_2^2... + a_{n-1}^2}}\right)^2$$ by Cauchy-Schwarz inequality. WLOG, take $a_1^2 + a_2^2 + ... + a_n^2 = 1$ (because of homogenity). Thus it remains to prove that $$\frac{a_1}{\sqrt{1-a_1^2}} + \frac{a_2}{\sqrt{1-a_2^2}}+ \ ... \ +\frac{a_n}{\sqrt{1-a_n^2}} \ge 2$$Now take $x_m = \frac{a_m}{\sqrt{1-a_m^2}} \ \forall \ 1 \le m \le n$. Then it remains to prove that $$x_1 + x_2 + ... + x_n \ge 2$$ Now since $a_m^2 =\frac{x_m^2}{1+x_m^2}$, we have $$\frac{x_1^2}{1+x_1^2} + \ ... \ + \frac{x_n^2}{1+x_n^2} = 1$$ But $\frac{x_m^2}{1+x_m^2} = x_m \left( \frac{x_m}{1+x_m^2} \right) \le x_m \left( \frac 12\right)$ since $x_m \ge 0$ and it follows that $$1 = x_1 \left( \frac{x_1}{1+x_1^2} \right) + ... + x_n \left( \frac{x_n}{1+x_n^2} \right) \le \frac{x_1}2 + ... + \frac{x_n}2 = \frac 12 \left(x_1 + ... + x_n \right)$$  i.e. $$x_1 + x_2 + ... + x_n \ge 2$$ Done.
Note that $2$ is the minimum attainable value of $$\frac{a_1}{\sqrt{1-a_1^2}} + \frac{a_2}{\sqrt{1-a_2^2}}+ \ ... \ +\frac{a_n}{\sqrt{1-a_n^2}} \ge 2$$ (take $a_1 = a_2 =\frac 1{\sqrt 2}$ and all other $a_i $s equal to $0$).
