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[不等式] existe one prove with AM-GM?

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tommywong Posted 2021-4-2 18:11 |Read mode
MihaiT:

thanks very much,but for this existe one prove with AM-GM?
prove for all x,y,z>0
$\displaystyle \sum_{cyc}(x^5y+2x^4y^2+x^4yz+x^3y^3+14x^2y^2z^2-9x^3y^2z-10x^3yz^2)\geq0$

THANKS VERY MUCH
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kuing Posted 2021-4-3 17:25
原题是啥?

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 Author| tommywong Posted 2021-4-20 18:29
回复 2# kuing

MihaiT:

Original pro blem it is from ARQADY Let $a$, $b$ and $c$ be positive numbers such that $a^2+b^2+c^2+abc=4.$ Prove that:
$$\displaystyle\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+8abc\geq11.$$AFTER CLASSICALL SUBSTITUTION:

$\displaystyle a=2\sqrt{\frac{yz}{(x+y)(x+z)}}$ , $ b=...$ , $c=...$ we have

$\displaystyle\sum_{cyc}\frac{y(y+z)}{x(x+z)}+\frac{64xyz}{(x+y)(y+z)(z+x)}\geq{11}$

or

$\displaystyle\sum_{cyc}(x^5y+2x^4y^2+x^4yz+x^3y^3+14x^2y^2z^2-9x^3y^2z-10x^3yz^2)\geq0$

but is very ugli now!

thanks if you have one idee

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