Prove

**dhiab** · August 13th 2009, 01:57 AM

$a_1 ,a_2 ,a_3 <br /> <br />$

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

**TheAbstractionist** · August 13th 2009, 05:23 AM

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

Hence

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

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

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

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

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