inequality

Apr 2010
456
1
prove that:

\(\displaystyle (a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}\)

I tried by expanding the square
 
Last edited:
Mar 2015
182
68
Universe 2.71828i3.14159
prove that:

\(\displaystyle (a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}\)

I tried by expanding the square
By expanding we get $2a+2b+2c \ge 2 \sqrt{ab} + 2 \sqrt{bc} + 2 \sqrt{ca} \Rightarrow a+b+c \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$.

$a+b+c = \dfrac{a+b}{2} + \dfrac{b+c}{2} + \dfrac{c+a}{2}$.

Now prove, $\dfrac{a+b}{2} \ge \sqrt{ab}$
 
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