# Problem 230 (van khea)

គេអោយ $a, b, c$ ជាបីចំនួនពិតវិជ្ជមានផ្ទៀងផ្ទាត់ $ab+bc+ca=1$ ។ ស្រាយបញ្ជាក់ថាៈ
$\displaystyle \frac{a}{(b+c)^3}+\frac{b}{(c+a)^3}+\frac{c}{(a+b)^3}\geq \frac{9}{8}$
សំរាយបញ្ជាក់
យើងមានៈ
$\displaystyle \frac{a}{(b+c)^3}+\frac{b}{(c+a)^3}+\frac{c}{(a+b)^3}$$\displaystyle =\frac{a(b+c)}{(b+c)^4}+\frac{b(c+a)}{(c+a)^4}+\frac{c(a+b)}{(a+b)^4}$
ឧបមាថា $\displaystyle a\leq b\leq c\Rightarrow \frac{1}{(a+b)^4}\geq \frac{1}{(a+b)^4}\geq \frac{1}{(a+b)^4}$$\& c(a+b)\geq b(c+a)\geq a(b+c)$
តាមវិសមភាព $Chebyshev$ យើងបានៈ
$\displaystyle \frac{a(b+c)}{(b+c)^4}+\frac{b(c+a)}{(c+a)^4}+\frac{c(a+b)}{(a+b)^4}$$\displaystyle \geq \frac{1}{3}(2(ab+bc+ca))(\frac{1}{(a+b)^4}+\frac{1}{(b+c)^4}+\frac{1}{(c+a)^4})$
$\displaystyle \Leftrightarrow \frac{a}{(b+c)^3}+\frac{b}{(c+a)^3}+\frac{c}{(a+b)^3}$$\displaystyle \geq \frac{2}{3}(\frac{1}{(a+b)^4}+\frac{1}{(b+c)^4}+\frac{1}{(c+a)^4})$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle \frac{1}{(a+b)^4}+\frac{1}{(b+c)^4}+\frac{1}{(c+a)^4}$$\displaystyle \geq \frac{1}{3}(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2})^2$
ម្យ៉ាងទៀតតាមវិសមភាព $Iran 1996$ យើងមានៈ
$\displaystyle \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\geq \frac{9}{4(ab+bc+ca)}=\frac{9}{4}$
ដូចនេះយើងទាញបានៈ $\displaystyle \frac{a}{(b+c)^3}+\frac{b}{(c+a)^3}+\frac{c}{(a+b)^3}\geq \frac{9}{8}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$