# Problem 251 (van khea)

គេអោយបីចំនួនពិតវិជ្ជមានផ្ទៀងផ្ទាត់ $a\leq b\leq c$ និង $a+b+c=3$ ។ ស្រាយបញ្ជាក់ថាៈ
$\displaystyle \sqrt[4]{a(a+b)}+\sqrt[4]{b(b+c)}+\sqrt[4]{c(c+a)}\geq \frac{\sqrt[4]{2}}{3}(ab+bc+ca)^2$
សំរាយបញ្ជាក់
យើងមានៈ
$\displaystyle \sqrt[4]{a(a+b)}=\frac{\sqrt[4]{a}}{\frac{1}{\sqrt[4]{a+b}}}=\frac{a}{\frac{\sqrt[4]{a^3}}{\sqrt[4]{a+b}}}=\frac{(\sqrt{a})^2}{\frac{\sqrt[4]{a^3}}{\sqrt[4]{a+b}}}$
ដូចនេះយើងទាញបានៈ
$\displaystyle \sqrt[4]{a(a+b)}+\sqrt[4]{b(b+c)}+\sqrt[4]{c(c+a)}$$\displaystyle =\frac{(\sqrt{a})^2}{\frac{\sqrt[4]{a^3}}{\sqrt[4]{a+b}}}+\frac{(\sqrt{b})^2}{\frac{\sqrt[4]{b^3}}{\sqrt[4]{b+c}}}+\frac{(\sqrt{c})^2}{\frac{\sqrt[4]{c^3}}{\sqrt[4]{c+a}}}$$\displaystyle \geq \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\frac{\sqrt[4]{a^3}}{\sqrt[4]{a+b}}+\frac{\sqrt[4]{b^3}}{\sqrt[4]{b+c}}+\frac{\sqrt[4]{c^3}}{\sqrt[4]{c+a}}}$
តាង $\displaystyle A=\frac{\sqrt[4]{a^3}}{\sqrt[4]{a+b}}+\frac{\sqrt[4]{b^3}}{\sqrt[4]{b+c}}+\frac{\sqrt[4]{c^3}}{\sqrt[4]{c+a}}$
យើងនឹងស្រាយថាចំពោះចំនួនពិតវិជ្ជមាន $\displaystyle a\leq b\leq c\Rightarrow A\leq \frac{3}{\sqrt[4]{2}}$
យើងមាន $\displaystyle A=\frac{\sqrt[4]{a^3}}{\sqrt[4]{a+b}}+\frac{\sqrt[4]{b^3}}{\sqrt[4]{b+c}}+\frac{\sqrt[4]{c^3}}{\sqrt[4]{c+a}}=\biggl(\frac{a}{\sqrt[3]{a+b}}\biggl)^{\frac{3}{4}}+\biggl(\frac{b}{\sqrt[3]{b+c}}\biggl)^{\frac{3}{4}}+\biggl(\frac{c}{\sqrt[3]{c+a}}\biggl)^{\frac{3}{4}}$$\displaystyle \leq \sqrt[4]{3}\biggl(\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\biggl)^{\frac{3}{4}}$
តាង $\displaystyle B=\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}$ យើងនឹងស្រាយថាចំពោះ $\displaystyle a\leq b\leq c\Rightarrow B\leq \frac{3}{\sqrt[3]{2}}$
តាង $x^3=a+b;y^3=c+a;z^3=b+c$
ដោយ $a\leq b\leq c\Rightarrow x\leq y\leq z$
យើងទាញបានៈ
$x^3+y^3-z^3=2a; z^3+x^3-y^3=2b;y^3+z^3-x^3=2c$
ដូចនេះយើងបានៈ
$\displaystyle B=\frac{(x^3+y^3-z^3)}{2x}+\frac{(y^3+z^3-x^3)}{2y}+\frac{(z^3+x^3-y^3)}{2z}$
តាមលំហាត់ទី 74 van khea ចំពោះ $x\leq y\leq z$ យើងមានៈ
$\displaystyle \frac{(x^3+y^3-z^3)}{x}+\frac{(y^3+z^3-x^3)}{y}+\frac{(z^3+x^3-y^3)}{z}\leq x^2+y^2+z^2$
$\displaystyle \Rightarrow B\leq \frac{1}{2}(x^2+y^2+z^2)$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle x^2+y^2+z^2=(a+b)^{\frac{2}{3}}+(b+c)^{\frac{2}{3}}+(c+a)^{\frac{2}{3}}$$\displaystyle \leq \sqrt[3]{3}(2(a+b+c)^{\frac{2}{3}}=3.2^{\frac{2}{3}}$
$\displaystyle \Rightarrow B\leq \frac{3}{\sqrt[3]{2}}$
$\displaystyle \Rightarrow A\leq \sqrt[4]{3}(\frac{3}{\sqrt[3]{2}})^{\frac{3}{4}}=\frac{3}{\sqrt[4]{2}}$
ដូចនេះយើងបានៈ
$\displaystyle \sqrt[4]{a(a+b)}+\sqrt[4]{b(b+c)}+\sqrt[4]{c(c+a)}\geq \frac{\sqrt[4]{2}}{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
ឥឡូវយើងនឹងស្រាយថាចំពោះចំនួនពិតវិជ្ជមាន $a, b, c$ យើងនឹងបានៈ $\sqrt{a}+\sqrt{a}+\sqrt{a}\geq ab+bc+ca$ ចំពោះគ្រប់ $a+b+c=3$
យើងមានៈ $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$\Rightarrow 2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2) ;(1)$
ម្យ៉ាងទៀតតាមវិសមភាព $AM-GM$ យើងមានៈ
$a^2+\sqrt{a}+\sqrt{a}\geq 3a$
$b^2+\sqrt{b}+\sqrt{b}\geq 3b$
$c^2+\sqrt{c}+\sqrt{c}\geq 3c$
យើងបានៈ
$a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 3(a+b+c) ; (2)$
បូក $(1)+(2)$ យើងបានៈ
$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ca$
ដូចនេះយើងបានៈ
$\displaystyle \sqrt[4]{a(a+b)}+\sqrt[4]{b(b+c)}+\sqrt[4]{c(c+a)}\geq \frac{\sqrt[4]{2}}{3}(ab+bc+ca)^2$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c=1$