Inequality

**banku12** · August 6th 2010, 08:34 PM

if $a,b,c,d>0$ and

$a+b+c+d=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{ 1}{d}$

prove that

$2(a+b+c+d)\geq\sqrt{a^{2}+3}+\sqrt{b^{2}+3}+\sqrt{ c^{2}+3}+\sqrt{d^{2}+3}$

**simplependulum** · August 6th 2010, 08:47 PM

Using Cachy Inequality , we have

$\sqrt{ \left( \sqrt{a}^2 + \sqrt{b}^2 + \sqrt{c}^2 + \sqrt{d}^2 \right) \left( \sqrt{ \frac{a^2+3}{a} }^2 + \sqrt{ \frac{b^2+3}{b} }^2 + \sqrt{ \frac{c^2+3}{c} }^2 + \sqrt{ \frac{d^2+3}{d} }^2 \right) } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}$

$\sqrt{ ( a+b+c+d) [ a + b + c + d + 3( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) ] } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}$

$\sqrt{ ( a+b+c+d) [ a + b + c + d + 3(a+b+c+d ) ] } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}$

$\sqrt{4 (a+b+c+d)^2 } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}$

$2(a+b+c+d) \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}$

**banku12** · August 6th 2010, 09:29 PM

Thank u

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