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For $ a,b,c,d\geq 0 $ with $ a+b = c+d = 2 $, how to prove that $$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$$

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    For starters, if $x,y\geq 0$, then $$(x+y)^2\geq x^2+y^2$$2012-08-24
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    @PeterTamaroff if you were thinking about changing all of the expressions above like you described, that wouldn't work since after taking the square root of both sides, and making $a=2,b=0,c=1,$and $d=1$, you get $9\le5$2012-08-24
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    @Sidd I don't understand what you mean. I just hinted a very obvious inequality. I don't know if it will work to prove this, and I actually scanned too fast for I thought there was an $a^2+b^2$ up there.2012-08-24
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    Peter's inequality looks familiar. Is there a name for that inequality2012-08-24
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    Just using $x\cdot y\geq0$ if $x,y\geq 0$. Expand the binomial. Obvious inequality is obvious =).2012-08-24
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    @PeterTamaroff. I am not askig how to get the inequality from the result, but the other way around. Did you build that from scratch? It resembles alot like the triangle inequality2012-08-24
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    The triangle inequality for the absolute value is $|x+y|\leq |x|+|y|$, and in general, for norms, it is $$\left\| {x+y} \right\|\leq \left\| {x} \right\|+\left\| {y} \right\|$$ Again, the above inequality is just a restatement if $x\cdot y \geq 0$ if $x,y\geq 0$. Just multiply by $2$ and sum $x^2+y^2$ to both sides.2012-08-24
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    I posted a [related question](http://math.stackexchange.com/questions/186557/bounding-the-product-of-a-quadrilaterals-side-lengths-in-terms-of-the-lengths-of).2012-08-25

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