# Problem 246 (van khea)

ស្រាយបញ្ជាក់ថាចំពោះបណ្ដាចំនួនពិតវិជ្ជមាន $a, b, c$ គេបានៈ
$\displaystyle \sum_{cyc}\frac{(a-b)^2}{a\sqrt{bc}}\geq 2\sum_{cyc}\frac{a-b}{a+b}.\frac{a-c}{b}+2\sum_{cyc}\frac{a-c}{a+c}.\frac{a-b}{c}$
សំរាយបញ្ជាក់
តាមវិសមភាព $AM-GM$ យើងមានៈ
$\displaystyle \sum_{cyc}\frac{(a-b)^2}{a\sqrt{bc}}\geq 2\sum_{cyc}\frac{(a-b)^2}{a(b+c)}$
ដូចនេះយើងត្រូវស្រាយថាៈ
$\displaystyle \sum_{cyc}\frac{(a-b)^2}{a(b+c)}\geq \sum_{cyc}\frac{a-b}{a+b}.\frac{a-c}{b}+\sum_{cyc}\frac{a-c}{a+c}.\frac{a-b}{c}$
តាង $\displaystyle f(x)=\frac{1}{x}$ និង $\displaystyle g(x)=\frac{1}{a+b+c-x}$
យើងបានៈ $\displaystyle f''(x)=\frac{2}{x^2}$ និង $\displaystyle g''(x)=\frac{2}{(a+b+c-x)^2}$
$\displaystyle \Rightarrow f''(x).g''(x)=\frac{4}{x^2(a+b+c-x)^2}>0$
តាង $\displaystyle h(x)=f(x).g(x)=\frac{1}{x(a+b+c-x)}$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle \Delta_a=\begin{vmatrix}f(b)&-f(c)\\g(b)&g(c)\end{vmatrix}=\begin{vmatrix}\frac{1}{b}&-\frac{1}{c}\\\frac{1}{c+a}&\frac{1}{a+b}\end{vmatrix}=\frac{1}{b(a+b)}+\frac{1}{c(c+a)}$
$\displaystyle \Delta_b=\begin{vmatrix}f(c)&-f(a)\\g(c)&g(a)\end{vmatrix}=\begin{vmatrix}\frac{1}{c}&-\frac{1}{a}\\\frac{1}{a+b}&\frac{1}{b+c}\end{vmatrix}=\frac{1}{c(c+b)}+\frac{1}{a(a+b)}$
$\displaystyle \Delta_c=\begin{vmatrix}f(a)&-f(b)\\g(a)&g(b)\end{vmatrix}=\begin{vmatrix}\frac{1}{a}&-\frac{1}{b}\\\frac{1}{b+c}&\frac{1}{c+a}\end{vmatrix}=\frac{1}{a(c+a)}+\frac{1}{b(b+c)}$
តាមវិសមភាព $h-\Delta$ $Van Khea$ យើងបានៈ
$\displaystyle \sum_{cyc}(a-b)^2h(c)\geq \sum_{cyc}(a-b)(a-c)\Delta_a$
$\displaystyle \Leftrightarrow \sum_{cyc}\frac{(a-b)^2}{a(b+c)}\geq \sum_{cyc}\frac{a-b}{a+b}.\frac{a-c}{b}+\sum_{cyc}\frac{a-c}{a+c}.\frac{a-b}{c}$
ដូចនេះយើងបានៈ
$\displaystyle \sum_{cyc}\frac{(a-b)^2}{a\sqrt{bc}}\geq 2\sum_{cyc}\frac{a-b}{a+b}.\frac{a-c}{b}+2\sum_{cyc}\frac{a-c}{a+c}.\frac{a-b}{c}$
សមភាពកើតមានពេល $a=b=c$