[不等式] 来自群的简单三元不等式

(117.71 KB)

2012-10-27 22:33

let $a=y/x$, $b=z/y$, $c=x/z$, $x$, $y$, $z>0$, using CS, Vasc's inequality and AG, we have
\begin{align*}
\sum\frac1{a(a+1)+ab(ab+1)}&=\sum\frac{x^2}{xy+y^2+zx+z^2}\\
&\geqslant \frac{\left(\sum x^2\right)^2}{\sum x^2(xy+y^2+zx+z^2)}\\
&=\frac{\left(\sum x^2\right)^2}{\sum x^3y+\sum xy^3+2\sum x^2y^2}\\
&\geqslant \frac{\left(\sum x^2\right)^2}{\frac13\left(\sum x^2\right)^2+\frac13\left(\sum x^2\right)^2+\frac23\left(\sum x^2\right)^2}\\
&=\frac34.
\end{align*}

不用 Vasc 不等式也行，配方易得
\[4\left(\sum x^2\right)^2-3\left(\sum x^3y+\sum xy^3+2\sum x^2y^2\right)=\sum(x^2+y^2-xy)(x-y)^2+\sum(x^2-y^2)^2.\]

2# kuing


忒强了吧，kk.我想不到这么多。