# Problem 245 (vankhea)

ស្រាយបញ្ជាក់ថាចំពោះបណ្ដាចំនួនពិតវិជ្ជមាន $a, b, c$ គេបានៈ
$\displaystyle \sum_{cyc}(a-b)^2c\geq 2\sum_{cyc}(a-b)(a-c)\sqrt{bc}$
សំរាយបញ្ជាក់
តាង $f(x)=g(x)=\sqrt{x}, x\geq 0$
$\displaystyle \Rightarrow f''(x)=g''(x)=-\frac{1}{4}x^{-\frac{3}{2}}$
$\displaystyle \Rightarrow f''(x).g''(x)=\frac{1}{16}x^{-3}>0$
តាង $h(x)=f(x).g(x)=(f.g)(x)=\sqrt{x}.\sqrt{x}=x>0$
យើងមានៈ
$\displaystyle \Delta_a=\begin{vmatrix}f(b)&-f(c)\\g(b)&g(c)\end{vmatrix}=\begin{vmatrix}\sqrt{b}&-\sqrt{c}\\\sqrt{b}&\sqrt{c}\end{vmatrix}=2\sqrt{bc}$
$\displaystyle \Delta_b=\begin{vmatrix}f(c)&-f(a)\\g(c)&g(a)\end{vmatrix}=\begin{vmatrix}\sqrt{c}&-\sqrt{a}\\\sqrt{c}&\sqrt{a}\end{vmatrix}=2\sqrt{ca}$
$\displaystyle \Delta_c=\begin{vmatrix}f(a)&-f(b)\\g(a)&g(b)\end{vmatrix}=\begin{vmatrix}\sqrt{a}&-\sqrt{b}\\\sqrt{a}&\sqrt{b}\end{vmatrix}=2\sqrt{ab}$
តាមវិសមភាព $h-\Delta$ $Van Khea$ យើងបានៈ
$\displaystyle (a-b)^2h(c)+(b-c)^2h(a)+(c-a)^2h(b)$$\geq (a-b)(a-c)\Delta_a+(b-a)(b-c)\Delta_b+(c-a)(c-b)\Delta_c$
$\displaystyle \Leftrightarrow \sum_{cyc}(a-b)^2c\geq 2\sum_{cyc}(a-b)(a-c)\sqrt{bc}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$