Inequality

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Inequality

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#1 h Nov 18, 2010, 9:28 PM • 2 Y

Y by Adventure10, Mango247

Given $a,b,c>0$ .Prove that:
$\frac {a+b+c}{3}- \sqrt[3]{abc} \le max$ { $(\sqrt {a}- \sqrt{b})^{2},(\sqrt {b}- \sqrt{c})^{2},(\sqrt {c}- \sqrt{a})^{2}$ }

This post has been edited 1 time. Last edited by EhsanNamdari, Nov 19, 2010, 6:51 AM

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#2 h Nov 18, 2010, 9:57 PM • 1 Y

Y by Adventure10

Is there $\sqrt[3]{abc}$ instead of $\sqrt[3]{xyz}$ ?

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#3 h Nov 19, 2010, 5:31 AM • 2 Y

Y by Adventure10, Mango247

EhsanNamdari wrote:

Given $a,b,c>0$ .Prove that:
$\frac {a+b+c}{3}- \sqrt[3]{abc} \le max\{ (\sqrt {a}- \sqrt{b})^{2},(\sqrt {b}- \sqrt{c})^{2},(\sqrt {c}- \sqrt{a})^{2}\}$

See here http://mathifc.wordpress.com/2010/10/28/inequality-41/

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#4 h Nov 19, 2010, 5:45 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I will prove the stronger one:
let $a,b,c\ge0$ , then have
$\frac {a+b+c}{3}- \sqrt[3]{abc} \le \frac{2}{3}\max\left\{ (\sqrt {a}- \sqrt{b})^{2},(\sqrt {b}- \sqrt{c})^{2},(\sqrt {c}- \sqrt{a})^{2}\right\}$

prove: let $a = x^6 ,b = y^6 ,c = z^6$ and WLOG assume $x \ge y \ge z \ge 0$ , then ineq becomes
$\frac{{x^6 + y^6 + z^6 }}{3} - x^2 y^2 z^2 \le \frac{2}{3}\left( {x^3 - z^3 } \right)^2$
or
$\left( {x^2 + y^2 + z^2 } \right)\left( {\left( {x^2 - y^2 } \right)^2 + \left( {y^2 - z^2 } \right)^2 + \left( {z^2 - x^2 } \right)^2 } \right) \le 4\left( {x^3 - z^3 } \right)^2 .$
because $x \ge y \ge z \ge 0$ we have $y^2\le x^2$ and $\left( {x^2 - y^2 } \right)^2 + \left( {y^2 - z^2 } \right)^2 \le \left( {x^2 - z^2 } \right)^2$ , so we just need to prove
$\left( {2x^2 + z^2 } \right)\left( {x^2 - z^2 } \right)^2 \le 2\left( {x^3 - z^3 } \right)^2$
or
$\left( {2x^2 + z^2 } \right)\left( {x + z} \right)^2 \left( {x - z} \right)^2 \le 2\left( {x^2 + xz + z^2 } \right)^2 \left( {x - z} \right)^2 .$
but
$\left( {2x^2 + z^2 } \right)\left( {x + z} \right)^2 \le 2\left( {x^2 + xz + z^2 } \right)^2 \iff z^2 \left(3 x^2+2 x z+z^2\right) \ge 0$
is obvious true, so the ineq get proved.

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