Inequality with $x^2+y^2+z^2+xyz=4$ condition

Asked Sep 21 '15 at 02:32

Active Sep 21 '15 at 02:41

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For $x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove that $$ 4(xy+yz+zx-xyz) \geqslant (x^2y+z)(y^2z+x)(z^2x+y)$$

Observations

The condition $x^2+y^2+z^2+xyz=4$ is special. One can use the transformation $x=2cosA,y=2cosB,z=2cosC$ and the identity $$cos^2A+cos^2B+cos^2C+2cosAcosBcosC =1$$ to solve the equivalent trig inequality. I attempted to go this road and used Blandon inequality, but it was too much computation.
The condition reminds us about the USAMO 2001 inequality. USAMO2001 stated that $$2\geqslant xy+yz+zx-xyz$$ sadly the sign is reversed.
This is a cyclic inequality. I don't think the trick of rewrite the inequality to $xyz \text{ }$,$ xy+yz+zx\text{ }$,$\text{ }x+y+z \text{ }$ worked here.

edited Sep 21 '15 at 02:41

asked Sep 21 '15 at 02:32

HN_NH

Try finding a solution (like $x=y=z=1$) and check if it is a global solution and check derivatives – Squirtle Sep 21 '15 at 02:48
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here's a good one: figure out how to construct all solutions, in positive integers, of $x^2 + y^2 + z^2 = 3xyz$ – Will Jagy Sep 21 '15 at 02:52
try this :$a=\dfrac{x}{yz},b=\dfrac{y}{xz},c=\dfrac{z}{xy}$, I am not sure it works, but it can simplify the inequality. – chenbai Sep 21 '15 at 08:46
Look there: http://math.stackexchange.com/questions/1805719/prove-sqrt2xy-sqrt3yz-sqrt4zx-4/1847546#1847546 – Yuri Negometyanov Jul 18 '16 at 11:25
Maybe we may replace the RHS with a symmetric form. – River Li Jun 26 '19 at 16:29

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