<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] 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 $0$).
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