三元不等式(a/b)(4a-3b-c)^2

tommywong

已知 $a,b,c>0$ 求證

$\displaystyle\frac{a}{b}(4a-3b-c)^2+\frac{b}{c}(4b-3c-a)^2+\frac{c}{a}(4c-3a-b)^2\ge 20(a^2+b^2+c^2-ab-bc-ca)$

hbghlyj

artofproblemsolving.com/community/c4h2015518p14144988

artofproblemsolving.com/community/c6h1476960p8597461

w.l.o.g. $abc=1$
\begin{gathered}x_a=4a^2-4b^2+\frac{8}{a}-\frac{5}{b}-\frac{3}{c}\\x_b=4b^2-4c^2+\frac{8}{b}-\frac{5}{c}-\frac{3}{a}\\x_c=4c^2-4a^2+\frac{8}{c}-\frac{5}{a}-\frac{3}{b}\\ \sum \frac{a}{b}(4a-3b-c)^2-10\sum (a-b)^2=\frac{\sum a(2ax_b+2bx_a-bx_c-cx_a)^2+14\sum x_a^2}{4\sum a^2+17\sum \frac{1}{a}}\ge 0\end{gathered}