# Problem 301 Van Khea

គេអោយ $a, b, c$ ជាចំនួនពិតវិជ្ជមាន។ ស្រាយបញ្ជាក់ថាៈ
Let $a, b, c$ be positive real numbers. Prove that:
$\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\geq \frac{3}{2}$
Solution
We have
$\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}$$\displaystyle =\frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}$
Because $\displaystyle \frac{3}{2}-\frac{1}{2}=1$ then we have
$\displaystyle \frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}$$\displaystyle \geq \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}$
Thus, it suffices to show that
$\displaystyle \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}\geq \frac{3}{2}$
$\displaystyle \Leftrightarrow 27((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)\leq 4(a^2+b^2+c^2)^3$
Letting $a^2=x; b^2=y; c^2=z$ yields
$\displaystyle 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3$
From $Cauchy-Schwarz$ inequality we have
$\displaystyle ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2$$\leq (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)$
So we need to prove that
$\displaystyle (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)\leq \frac{16}{27^2}(x+y+z)^6$
Now we will show that for $x, y, z$ are positive real numbers then we have
$27(x^2y+y^2z+z^2x+xyz)\leq 4(x+y+z)^3$ ;$(1)$ and $27(xy^2+yz^2+zx^2+xyz)\leq 4(x+y+z)^3$ ; $(2)$
without loss of generality, suppose that $x=min(x, y, z)$. Sitting $y=x+u$ and $z=x+v; (u, v\geq 0)$ then:
$(1)\Leftrightarrow 9(u^2-uv+v^2)x+(u-2v)^2(4u+v)\geq 0$
$(2)\Leftrightarrow 9(u^2-uv+v^2)x+(2u-v)^2(u+4v)\geq 0$
which is obviously true. Equality occurs for $u=v=0\Leftrightarrow x=y=z$
$\displaystyle \Rightarrow ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2$$\displaystyle \leq \frac{16}{27^2}(x+y+z)^6$
$\displaystyle \Leftrightarrow 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3$ is true.
Therefore the proof is completed. Equality occurs for $x=y=z\Leftrightarrow a=b=c$