Problem 232 (van khea)

ឧបមាថា $a, b, c$ ជាបីចំនួនពិតវិជ្ជមានផ្ទៀងផ្ទាត់ $abc(a^3+b^3+c^3)\leq 3$។ ចូរស្រាយបញ្ជាក់ថាៈ
$\displaystyle \frac{a^3}{b^2+c^4}+\frac{b^3}{c^2+a^4}+\frac{c^3}{a^2+b^4}\geq \frac{3\sqrt[3]{abc}}{1+\sqrt[3]{a^2b^2c^2}}$
សំរាយបញ្ជាក់
ដោយ $a, b, c>0$ នោះយើងអាចតាង $abc=k^3; k>0$ ដូចនេះយើងត្រូវស្រាយថាៈ
$\displaystyle \frac{a^3}{b^2+c^4}+\frac{b^3}{c^2+a^4}+\frac{c^3}{a^2+b^4}\geq \frac{3k}{1+k^2}$
យើងមានៈ $\displaystyle \frac{a^3}{b^2+c^4}+\frac{b^3}{c^2+a^4}+\frac{c^3}{a^2+b^4}$$\displaystyle =\frac{(a^3)^2}{a^3b^2+c^4a^3}+\frac{(b^3)^2}{b^3c^2+a^4b^3}+\frac{(c^3)^2}{c^3a^2+b^4c^3}$$\displaystyle \geq \frac{(a^3+b^3+c^3)^2}{a^3b^2+b^3c^2+c^3a^2+a^4b^3+b^4c^3+c^4a^3}$
យើងចាំបាច់ត្រូវស្រាយថាៈ
$(a^3+b^3+c^3)^{5}\geq 9(a^3b^2+b^3c^2+c^3a^2)^3$
$(a^3+b^3+c^3)^7\geq 81(a^4b^3+b^4c^3+c^4a^3)^3$
តាង $\displaystyle x=a^3; y=b^3; z=c^3$ នោះយើងត្រូវស្រាយថាៈ
$\displaystyle (x+y+z)^5\geq 9(xy^{\frac{2}{3}}+yz^{\frac{2}{3}}+zx^{\frac{2}{3}})^3$
$\displaystyle (x+y+z)^7\geq 81(x^{\frac{4}{3}}y+y^{\frac{4}{3}}z+z^{\frac{4}{3}}x)^3$
យើងមាន
$\displaystyle ((x^2y)^{\frac{1}{3}}.(xy)^{\frac{1}{3}}.1^{\frac{1}{3}}+(y^2z)^{\frac{1}{3}}.(yz)^{\frac{1}{3}}.1^{\frac{1}{3}}+(z^2x)^{\frac{1}{3}}.(zx)^{\frac{1}{3}}.1^{\frac{1}{3}})^3$$\displaystyle \leq 3(x^2y+y^2z+z^2x)(xy+yz+zx)$
$\displaystyle \Leftrightarrow (xy^{\frac{2}{3}}+yz^{\frac{2}{3}}+zx^{\frac{2}{3}})^3\leq 3(x^2y+y^2z+z^2x)(xy+yz+zx)$$\displaystyle \leq \frac{(x+y+z)^5}{9}$
$\displaystyle \Rightarrow (x+y+z)^5\geq 9(xy^{\frac{2}{3}}+yz^{\frac{2}{3}}+zx^{\frac{2}{3}})^3$
ដូចគ្នាដែរយើងមានៈ
$\displaystyle ((x^2y)^{\frac{1}{3}}(xy)^{\frac{1}{3}}(xy)^{\frac{1}{3}}+(x^2y)^{\frac{1}{3}}(xy)^{\frac{1}{3}}(xy)^{\frac{1}{3}}+(x^2y)^{\frac{1}{3}}(xy)^{\frac{1}{3}}(xy)^{\frac{1}{3}})^3$$\displaystyle \leq (x^2y+y^2z+z^2x)(xy+yz+zx)^2$$\displaystyle \leq \frac{(x+y+z)^5}{27}.\frac{(x+y+z)^2}{3}=\frac{(x+y+z)^7}{81}$
$\displaystyle \Rightarrow (x+y+z)^7\geq 81(x^{\frac{4}{3}}y+y^{\frac{4}{3}}z+z^{\frac{4}{3}}x)^3$
ដូចនេះវិសមភាពខាងលើសមមូលនឹងៈ
$\displaystyle \frac{a^3}{b^2+c^4}+\frac{b^3}{c^2+a^4}+\frac{c^3}{a^2+b^4}$$\displaystyle \geq \frac{(a^3+b^3+c^3)^2}{\frac{(a^3+b^3+c^3)^{\frac{5}{3}}}{\sqrt[3]{9}}+\frac{(a^3+b^3+c^3)^{\frac{7}{3}}}{3\sqrt[3]{3}}}$$\displaystyle \geq \frac{3\sqrt[3]{9}(a^3+b^3+c^3)^{\frac{1}{3}}}{3+\sqrt[3]{3}(a^3+b^3+c^3)^{\frac{2}{3}}}$
ដូចនេះយើងត្រូវស្រាយថាៈ $\displaystyle \frac{3\sqrt[3]{9}(a^3+b^3+c^3)^{\frac{1}{3}}}{3+\sqrt[3]{3}(a^3+b^3+c^3)^{\frac{2}{3}}}$$\displaystyle \geq \frac{3\sqrt[3]{abc}}{1+\sqrt[3]{a^2b^2c^2}}$
តាង $\displaystyle t^3=a^3+b^3+c^3\& k^3=abc$ នោះយើងបានៈ
$\displaystyle \frac{3\sqrt[3]{9}t}{3+\sqrt[3]{3}t^2}\geq \frac{3k}{1+k^2}$
$\displaystyle \Leftrightarrow (t-\sqrt[3]{3}k)(\sqrt[3]{9}-\sqrt[3]{3}kt)\geq 0$ ពិតជានិច្ចតាមលក្ខខណ្ឌលំហាត់ខាងលើ។
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជា់។ សមភាពកើតមានពេល $a=b=c\leq 1$