# Problem 269 van khea

ស្រាយបញ្ជាក់ថាចំពោះបណ្ដាចំនួនពិតវិជ្ជមាន $a, b, c$ គេបានៈ
$\displaystyle a^2\sqrt{1+\frac{b}{c}}+b^2\sqrt{1+\frac{c}{a}}+c^2\sqrt{1+\frac{a}{b}}\geq \frac{\sqrt{2}}{3}(a+b+c)^2$
សំរាយបញ្ជាក់
យើងមានៈ
$\displaystyle a^2\sqrt{1+\frac{b}{c}}+b^2\sqrt{1+\frac{c}{a}}+c^2\sqrt{1+\frac{a}{b}}$$\displaystyle =\frac{a^2}{\sqrt{\frac{c}{b+c}}}+\frac{b^2}{\sqrt{\frac{a}{c+a}}}+\frac{c^2}{\sqrt{\frac{b}{a+b}}}$
$\displaystyle \geq \frac{(a+b+c)^2}{\sqrt{\frac{a}{c+a}}+\sqrt{\frac{b}{a+b}}+\sqrt{\frac{c}{b+c}}}$
ដូចនេះយើងត្រូវស្រាយថាៈ $\displaystyle \sqrt{\frac{a}{c+a}}+\sqrt{\frac{b}{a+b}}+\sqrt{\frac{c}{b+c}}\leq \frac{3}{\sqrt{2}}$
$\displaystyle \Leftrightarrow \biggl(\sqrt{\frac{2a}{c+a}}+\sqrt{\frac{2b}{a+b}}+\sqrt{\frac{2c}{b+c}}\biggl)^2\leq 9$
តាង $\displaystyle P^2=\biggl(\sqrt{\frac{2a}{c+a}}+\sqrt{\frac{2b}{a+b}}+\sqrt{\frac{2c}{b+c}}\biggl)^2$
$\displaystyle P^2=\biggl(\sqrt{\frac{2a(a+b)}{(a+b)(a+c)}}+\sqrt{\frac{2b(b+c)}{(b+a)(b+c)}}+\sqrt{\frac{2c(c+a)}{(c+a)(c+b)}}\biggl)^2$
$\displaystyle \leq 2(a+b+c)\biggl(\frac{2a}{(a+b)(a+c)}+\frac{2b}{(b+a)(b+c)}+\frac{2c}{(c+a)(c+b)}\biggl)$$\leq 9$
$\displaystyle \Leftrightarrow 8(a+b+c)(ab+bc+ca)\leq 9(a+b)(b+c)(c+a)$
$\displaystyle \Leftrightarrow a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\geq 6abc$
តាមវិសមភាព $AM-GM$ យើងបានៈ
$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\geq 6\sqrt[6]{a^6b^6c^6}=6abc$ ពិត។
ដូចនេះវិសមភាពត្រូវស្រាយបញ្ចាក់។ សមភាពកើតមានពេល $a=b=c$