A new a, b, c - inequality / MB-166

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A new a, b, c - inequality / MB-166

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#1 h Oct 30, 2010, 9:56 AM • 2 Y

Y by Adventure10, Mango247

For $a, b, c >0,$ prove that the following inequality holds
$\frac{2a+3b+3c}{\sqrt{6a^2+5b^2+5c^2}}+\frac{3a+2b+3c}{\sqrt{5a^2+6b^2+5c^2}}+\frac{3a+3b+2c}{\sqrt{5a^2+5b^2+6c^2}}\leq 6$

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kuing

#2 h Oct 30, 2010, 11:16 AM • 2 Y

Y by Adventure10, Mango247

marin.bancos wrote:

For $a, b, c >0,$ prove that the following inequality holds
$\frac{2a+3b+3c}{\sqrt{6a^2+5b^2+5c^2}}+\frac{3a+2b+3c}{\sqrt{5a^2+6b^2+5c^2}}+\frac{3a+3b+2c}{\sqrt{5a^2+5b^2+6c^2}}\leq 6$

from $b^2+c^2\ge\frac{1}{2}(b+c)^2$ an so on, we know that just need to prove
$\sum {\frac{{2a + 3b + 3c}}{{\sqrt {6a^2 + \frac{5}{2}\left( {b + c} \right)^2 } }}} \le 6$
WLOG assume $a+b+c=1$ , ineq becomes
$\sum {\frac{{3 - a}}{{\sqrt {6a^2 + \frac{5}{2}\left( {1 - a} \right)^2 } }}} \le 6$
but
$\left( {\frac{{19 - 9a}}{8}} \right)^2 - \frac{{\left( {3 - a} \right)^2 }}{{6a^2 + \frac{5}{2}\left( {1 - a} \right)^2 }} = \frac{{(3a - 1)^2 \left( {153a^2 + 634(1 - a) + 19} \right)}}{{64(17a^2 - 10a + 5)}} \ge 0$
gives
$\sum {\frac{{3 - a}}{{\sqrt {6a^2 + \frac{5}{2}\left( {1 - a} \right)^2 } }}} \le \sum {\frac{{19 - 9a}}{8}} = 6.$
done.

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#3 h Oct 30, 2010, 11:56 AM • 2 Y

Y by Adventure10, Mango247

HI!
we know that $(b+c)^2 \leq 2(b^2+c^2)$
so It suffices to show that :
$\sum \frac{2a+3\sqrt{2(b^2+c^2)}}{\sqrt{5(a^2+b^2+c^2)+a^2}} \leq 6$
WLOG assume $a^2+b^2+c^2=1$
the inequality becomes :^
$\sum \frac{2a+3\sqrt{2-2a^2}}{\sqrt{5+a^2}} \leq 6$
which is true since
$\sum \sqrt{\frac{a^2}{5+a^2}} \leq \frac{3}{4}$
and
$\sum \sqrt{\frac{2-2a^2}{5+a^2}} \leq \frac{3}{2}$

This post has been edited 3 times. Last edited by Sporovitch, Oct 30, 2010, 4:21 PM

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kuing

#4 h Oct 30, 2010, 12:09 PM • 2 Y

Y by Adventure10, Mango247

Sporovitch wrote:

Hi!
WLOG assume a+b+c=1
According to CS
$(6a^2+5b^2+5c^2)(6+5+5) \geq (6a+5a+5b)^2=(5+a)^2$
so $\sum \frac{\frac{5+a}{2}}{\sqrt{6a^2+5b^2+5c^2}} \leq 6$
Thus it suffices to show that :
$\sum \frac{\frac{5+a}{2}}{\sqrt{6a^2+5b^2+5c^2}} \geq \sum \frac{3-a}{\sqrt{6a^2+5b^2+5c^2}}$
which is equivalent to :
$\sum \frac{3a-1}{\sqrt{6a^2+5b^2+5c^2}} \geq 0$
which is obviously true since
$\frac{1}{\sqrt{6a^2^+5b^2+5c^2}} > \frac{1}{\sqrt{6a^2+6b^2+6c^2}} $
Done !

$\sum \frac{3a-1}{\sqrt{6a^2+5b^2+5c^2}} \geq 0$
dose not hold, try $a=b\to0,c\to1$ .
in fact, I think $\sum \frac{3a-1}{\sqrt{6a^2+5b^2+5c^2}} \leq 0$

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#5 h Oct 30, 2010, 12:21 PM • 2 Y

Y by Adventure10, Mango247

Yes thank you dear kuing you're right
Sorry

its edited now

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#6 h Oct 30, 2010, 7:50 PM • 2 Y

Y by Adventure10, Mango247

Thank you very much for your solutions! They are different than mine.

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