# Problem 227 (van khea)

ស្រាយបញ្ជាក់ថាចំពោះបីចំនួនពិតវិជ្ជមាន $a, b, c, p, q$ គេបានៈ
$\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}$
សំរាយបញ្ជាក់
យើងមានៈ
$\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}$$\displaystyle =\frac{1}{a^{\frac{2}{3}}.a^{\frac{1}{3}}\sqrt[3]{pb+qc}}+\frac{1}{b^{\frac{2}{3}}.b^{\frac{1}{3}}\sqrt[3]{pc+qa}}+\frac{1}{c^{\frac{2}{3}}.c^{\frac{1}{3}}\sqrt[3]{pa+qb}}$$\displaystyle =\frac{1}{a^{\frac{2}{3}}\sqrt[3]{pab+qca}}+\frac{1}{b^{\frac{2}{3}}\sqrt[3]{pbc+qab}}+\frac{1}{c^{\frac{2}{3}}\sqrt[3]{pca+qbc}}$
តាង $\displaystyle abc=k\Rightarrow \frac{1}{a}=\frac{bc}{k};\frac{1}{b}=\frac{ca}{k};\frac{1}{c}=\frac{ab}{k}$
ដូចនេះយើងទាញបានថាៈ
$\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}$$\displaystyle =\frac{(bc)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pab+qca}}+\frac{(ca)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pbc+qab}}+\frac{(ab)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pca+qbc}}$
តាង $\displaystyle ab=x^3; bc=y^3; ca=z^3$ នោះយើងបានៈ
$\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}$$\displaystyle =\frac{1}{k^{\frac{2}{3}}}\biggl(\frac{x^2}{\sqrt[3]{pz^3+qy^3}}+\frac{y^2}{\sqrt[3]{px^3+qz^3}}+\frac{z^2}{\sqrt[3]{py^3+qx^3}}\biggl)$$\displaystyle =\frac{1}{k^{\frac{2}{3}}}\biggl(\frac{(x^2)^{\frac{4}{3}}}{\sqrt[3]{pz^3x^2+qx^2y^3}}+\frac{(y^2)^{\frac{4}{3}}}{\sqrt[3]{px^3y^2+qy^2z^3}}+\frac{(z^2)^{\frac{4}{3}}}{\sqrt[3]{py^3z^2+qz^2x^3}}\biggl)$$\displaystyle \geq \frac{(x^2+y^2+z^2)^{\frac{4}{3}}}{k^{\frac{2}{3}}.\sqrt[3]{p(x^3y^2+y^3z^2+z^3x^2)+q(x^2y^3+y^2z^3+z^2x^3)}}$
ម្យ៉ាងទៀតតាមលំហាត់ទី 32 vankhea យើងមានៈ
$\displaystyle (x^2+y^2+z^2)^{\frac{5}{2}}\geq 3\sqrt{3}(x^3y^2+y^3z^2+z^3x^2)$ និង $\displaystyle (x^2+y^2+z^2)^{\frac{5}{2}}\geq 3\sqrt{3}(x^2y^3+y^2z^3+z^2x^3)$
$\displaystyle \Rightarrow \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}$$\displaystyle \geq \frac{\sqrt{3}(x^2+y^2+z^2)^{\frac{4}{3}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}(x^2+y^2+z^2)^{\frac{5}{6}}}$$\displaystyle \geq \frac{\sqrt{3}(x^2+y^2+z^2)^{\frac{1}{2}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}}$
តាមវិសមភាព $AM-GM$ យើងមានៈ $\displaystyle x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}=3\sqrt[3]{(abc)^{\frac{4}{3}}}=3k^{\frac{4}{9}}$
$\displaystyle \Rightarrow \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3k^{\frac{2}{9}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}}=\frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$