<s>[INST] 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 .
 [/INST] 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.
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