An a,b,c,d - inequality ( abcd=1 ) / MB-139

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An a,b,c,d - inequality ( abcd=1 ) / MB-139

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#1 h Sep 28, 2010, 2:51 PM • 9 Y

Y by Adventure10, Mango247, and 7 other users

For $a, b, c, d>0,$ such that $abcd=1,$ prove that the following inequality holds
$\frac{1}{a^2(b+1)^2}+\frac{1}{b^2(c+1)^2}+\frac{1}{c^2(d+1)^2}+\frac{1}{d^2(a+1)^2}\geq1$

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kuing

#2 h Sep 28, 2010, 3:40 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

marin.bancos wrote:

For $a, b, c, d>0,$ such that $abcd=1,$ prove that the following inequality holds
$\frac{1}{a^2(b+1)^2}+\frac{1}{b^2(c+1)^2}+\frac{1}{c^2(d+1)^2}+\frac{1}{d^2(a+1)^2}\geq1$

$a = \frac{y}{x},b = \frac{z}{y},c = \frac{w}{z},d = \frac{x}{w},x,y,z,w > 0.$

then

$\frac{1}{{a^2 (b + 1)^2 }} + \frac{1}{{b^2 (c + 1)^2 }} + \frac{1}{{c^2 (d + 1)^2 }} + \frac{1}{{d^2 (a + 1)^2 }}$

$= \frac{{x^2 }}{{(y + z)^2 }} + \frac{{y^2 }}{{(w + z)^2 }} + \frac{{z^2 }}{{(x + w)^2 }} + \frac{{w^2 }}{{(x + y)^2 }}$

$\ge \frac{{(x^2 + y^2 + z^2 + w^2 )^2 }}{{x^2 (y + z)^2 + y^2 (z + w)^2 + z^2 (w + x)^2 + w^2 (x + y)^2 }}$

but

$(x^2 + y^2 + z^2 + w^2 )^2 - \left( {x^2 (y + z)^2 + y^2 (z + w)^2 + z^2 (w + x)^2 + w^2 (x + y)^2 } \right)$

$= (x^2 - yz)^2 + (y^2 - zw)^2 + (z^2 - wx)^2 + (w^2 - xy)^2 \ge 0$

done

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#3 h Sep 28, 2010, 3:56 PM • 2 Y

Y by Adventure10, Mango247

nice proof

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#4 h Sep 28, 2010, 4:45 PM • 2 Y

Y by Adventure10, Mango247

kuing's proof is very nice, indeed. And different than mine. Congratulations, dear kuing!
Just a remark ...
I can't understand why the problem was rated as spam. I really think that this inequality is nice. And its expression is beautiful.

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