# Problem 247 (van khea)

ឧបមាថា $a, b, c$ ជាបីចំនួនពិតវិជ្ជមានផ្ទៀងផ្ទាត់ $a\leq b\leq c$ និង $ab+bc+ca=1$ ។ ស្រាយបញ្ជាក់ថាៈ
$\displaystyle \frac{(ab)^{\frac{2}{3}}}{\sqrt{1+b^2}}+\frac{(bc)^{\frac{2}{3}}}{\sqrt{1+c^2}}+\frac{(ca)^{\frac{2}{3}}}{\sqrt{1+a^2}}\leq \sqrt[3]{\frac{9}{8}(a+b+c)}$
សំរាយបញ្ជាក់
យើងមាន $1+b^2=ab+bc+ca+b^2=(a+b)(b+c)$
$\displaystyle \Rightarrow \frac{(ab)^{\frac{2}{3}}}{\sqrt{1+b^2}}=\frac{a^{\frac{2}{3}}}{\sqrt{a+b}}.\frac{b^{\frac{2}{3}}}{\sqrt{b+c}}$
ដូចនេះយើងបានៈ
$\displaystyle \frac{(ab)^{\frac{2}{3}}}{\sqrt{1+b^2}}+\frac{(bc)^{\frac{2}{3}}}{\sqrt{1+c^2}}+\frac{(ca)^{\frac{2}{3}}}{\sqrt{1+a^2}}$$\displaystyle =\frac{a^{\frac{2}{3}}}{\sqrt{a+b}}.\frac{b^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{b^{\frac{2}{3}}}{\sqrt{b+c}}.\frac{c^{\frac{2}{3}}}{\sqrt{c+a}}+\frac{c^{\frac{2}{3}}}{\sqrt{c+a}}.\frac{a^{\frac{2}{3}}}{\sqrt{a+b}}$
$\displaystyle \leq \frac{1}{3}\biggl(\frac{a^{\frac{2}{3}}}{\sqrt{a+b}}+\frac{b^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{c^{\frac{2}{3}}}{\sqrt{c+a}}\biggl)^2$
ដូចនេះយើងត្រូវស្រាយថាៈ
$\displaystyle \frac{1}{3}\biggl(\frac{a^{\frac{2}{3}}}{\sqrt{a+b}}+\frac{b^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{c^{\frac{2}{3}}}{\sqrt{c+a}}\biggl)^2\leq \sqrt[3]{\frac{9}{8}(a+b+c)}$
$\displaystyle \Leftrightarrow \frac{a^{\frac{2}{3}}}{\sqrt{a+b}}+\frac{b^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{c^{\frac{2}{3}}}{\sqrt{c+a}}\leq \sqrt[6]{\frac{243}{8}(a+b+c)}$
តាង $\displaystyle x=\sqrt{a+b}; y=\sqrt{c+a}, z=\sqrt{b+c}$
តាមសម្មតិកម្មយើងមាន $a\leq b\leq c\Rightarrow x\leq y\leq z$
យើងទាញបានៈ
$x^2+y^2-z^2=2a$
$y^2+z^2-x^2=2c$
$z^2+x^2-y^2=2b$
ដូចនេះយើងត្រូវស្រាយថាៈ
$\displaystyle \frac{(x^2+y^2-z^2)^{\frac{2}{3}}}{x}+\frac{(z^2+x^2-y^2)^{\frac{2}{3}}}{z}+\frac{(y^2+z^2-x^2)^{\frac{2}{3}}}{y}\leq \sqrt[6]{486(a+b+c)}$
យើងមានៈ
$\displaystyle \frac{(x^2+y^2-z^2)^{\frac{2}{3}}}{x}+\frac{(z^2+x^2-y^2)^{\frac{2}{3}}}{z}+\frac{(y^2+z^2-x^2)^{\frac{2}{3}}}{y}$$\displaystyle \leq 3^{\frac{1}{3}}\biggl(\frac{x^2+y^2-z^2}{x^{\frac{3}{2}}}+\frac{y^2+z^2-x^2}{y^{\frac{3}{2}}}+\frac{z^2+x^2-y^2}{z^{\frac{3}{2}}}\biggl)^{\frac{2}{3}}$
តាមលំហាត់ទី 74 van khea ចំពោះចំនួនពិតវិជ្ជមាន $a\leq b\leq c$ និងចំពោះ $r, s\geq 0$ យើងមានៈ
$\displaystyle \frac{a^r+b^r-c^r}{a^s}+\frac{b^r+c^r-a^r}{b^s}+\frac{c^r+a^r-b^r}{c^s}\leq a^{r-s}+b^{r-s}+c^{r-s}$
ដូចនេះយើងបានៈ
$\displaystyle \frac{x^2+y^2-z^2}{x^{\frac{3}{2}}}+\frac{y^2+z^2-x^2}{y^{\frac{3}{2}}}+\frac{z^2+x^2-y^2}{z^{\frac{3}{2}}}\leq x^{\frac{1}{2}}+y^{\frac{1}{2}}+z^{\frac{1}{2}}$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle x^{\frac{1}{2}}+y^{\frac{1}{2}}+z^{\frac{1}{2}}=\sqrt[4]{a+b}+\sqrt[4]{b+c}+\sqrt[4]{c+a}$$\displaystyle \leq 3^{\frac{3}{4}}(2(a+b+c))^{\frac{1}{4}}$
$\displaystyle \Rightarrow \frac{(x^2+y^2-z^2)^{\frac{2}{3}}}{x}+\frac{(z^2+x^2-y^2)^{\frac{2}{3}}}{z}+\frac{(y^2+z^2-x^2)^{\frac{2}{3}}}{y}$$\displaystyle \leq 3^{\frac{1}{3}}(3^{\frac{3}{4}}(2(a+b+c))^{\frac{1}{4}})^{\frac{2}{3}}$$\leq \sqrt[6]{486(a+b+c)}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $\displaystyle a=b=c=\frac{\sqrt{3}}{3}$