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The positive real numbers $x,y,z$ are the side lengths of a triangle iff $$x^2 + y^2 + z^2 < 2\sqrt{x^2y^2 + y^2z^2 + z^2x^2}$$

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you have to replace $c$ with $z$. –  Iuli Feb 28 '13 at 20:45
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How can this post (which isn't even a question) have two upvotes? –  Git Gud Feb 28 '13 at 20:59
    
Yeah, this isn't a question... –  SSumner Feb 28 '13 at 21:01
    
Try put $x=a+b$,$y=b+c$,$z=c+a$. –  Yimin Feb 28 '13 at 21:44
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This is a perfectly valid question. Why do you say it isn't? The sides of a triangle need to satisfy the triangle inequality, and this inequality might be equivalent. –  robjohn Feb 28 '13 at 21:52

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up vote 6 down vote accepted

$$ (x^2+y^2+z^2)^2\lt4(x^2y^2+y^2z^2+z^2x^2)\\ \Updownarrow\\ x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2\lt0\\ \Updownarrow\\ (x-y)^2(x+y)^2+z^4-2z^2\left(x^2+y^2\right)\lt0\\ \Updownarrow\\ \left(z^2-(x-y)^2\right)\left(z^2-(x+y)^2\right)\lt0\\ \Updownarrow\\ (z-x+y)(z+x-y)(x+y-z)(z+x+y)\gt0 $$ Which is true if and only if the Triangle Inequality is true for all sides.

If the quantity is not positive, at least one of the sides fails.

The perimeter factor, $x+y+z$, must be positive.

If two of the "triangle" factors is not positive, then one of the sides is not positive; e.g. if $(z-x+y)\le0$ and $(z+x-y)\le0$, then $2z=(z-x+y)+(z+x-y)\le0$. Thus, only one of the triangle factors can be negative.

So if the quantity is positive, all the factors must be positive.


Heron's Formula says that the area of a triangle is $$ \frac14\sqrt{(z-x+y)(z+x-y)(x+y-z)(z+x+y)} $$ which is $$ \frac14\sqrt{4(x^2y^2+y^2z^2+z^2x^2)-(x^2+y^2+z^2)^2} $$

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