1. ## Prove

$\displaystyle a_1 ,a_2 ,a_3$

is the positifs reals numbers, Prove :
$\displaystyle 3\left( {\frac{1}{{a_1 a_2 }} + \frac{1}{{a_3 a_2 }} + \frac{1}{{a_1 a_3 }}} \right) \ge 4\left( {\frac{1}{{a_1 + a_2 }} + \frac{1}{{a_3 + a_2 }} + \frac{1}{{a_1 + a_3 }}} \right)^2$

2. First note that by AM–GM, $\displaystyle \frac{(a_i+a_j)^2}4\ge a_ia_j$ and so $\displaystyle \frac{a_ia_j}{(a_i+a_j)^2}\le\frac14.$

Hence

$\displaystyle \left(\frac{1}{{a_1 + a_2 }} + \frac{1}{{a_2 + a_3 }} + \frac{1}{{a_3 + a_1 }}\right)^2$

$\displaystyle =\ \left(\frac{\sqrt{a_1a_2}}{(a_1 + a_2)\sqrt{a_1a_2}} + \frac{\sqrt{a_2a_3}}{(a_2 + a_3)\sqrt{a_2a_3}} + \frac{\sqrt{a_3a_1}}{(a_3 + a_1)\sqrt{a_3a_1}}\right)^2$

$\displaystyle \leqslant\ \left(\frac{a_1a_2}{(a_1+a_2)^2}+\frac{a_2a_3}{(a_ 2+a_3)^2}+\frac{a_3a_1}{(a_3+a_1)^2}\right)\left(\ frac1{a_1a_2}+\frac1{a_2a_3}+\frac1{a_3a_1}\right)$ (Cauchy–Schwarz)

$\displaystyle \leqslant\ \frac34\left(\frac1{a_1a_2}+\frac1{a_2a_3}+\frac1{ a_3a_1}\right)$ (by above)