安振平老师的4901号不等式问题的证明

题目：已知$a,b,c>0$, $ab+bc+ca+abc=4$, 求证：$\frac{1}{a^2-a+1}+\frac{1}{b^2-b+1}+\frac{1}{c^2-c+1}\leq 3$.

证明：作代换$a=\frac{2x}{y+z}$, $b=\frac{2y}{z+x}$, $c=\frac{2z}{x+y}$, 则原不等式可化为

$\frac{(y+z)^2}{4x^2-2x(y+z)+(y+z)^2}+\frac{(z+x)^2}{4y^2-2y(z+x)+(z+x)^2}+\frac{(x+y)^2}{4z^2-2z(x+y)+(x+y)^2}\leq 3$.

即

$\sum{[9(y+z)x^3+(7y^2+7z^2-20yz)x^2-5(y^3+z^3)x+2yz(y^2+z^2)+y^4+z^4](y-z)^2}\geq 0$.

又

$9(y+z)x^3+(7y^2+7z^2-20yz)x^2-5(y^3+z^3)x+2yz(y^2+z^2)+y^4+z^4=\left[9t^3+7\left(t-\frac{5}{14}\right)^2+\frac{3}{28}\right](y-z)^4$

$+\left[45t^3+3t(3t-1)^2+40\left(t-\frac{7}{20}\right)^2+\frac{11}{10}\right]yz(y-z)^2+2\left[8t^3+4t(4t-1)^2+20\left(t-\frac{7}{20}\right)^2+\frac{11}{20}\right]y^2z^2\geq 0$

($其中t=\frac{x}{y+z}>0$)

于是原不等式成立.