$VP\geq \sqrt{\frac{9}{2}+\frac{3.6}{4}}=3$Vậy ta cm $VT\leq 3$
Thật vậy: $VT^2\leq 3(\frac{a+b}{c+ab}+\frac{b+c}{a+bc}+\frac{c+a}{b+ca})$
Ta có :$\frac{a+b}{c+ab}\leq \frac{a+b}{2\sqrt{cab}}\leq \frac{a+b}{2\sqrt{c}\frac{a+b}{2}}=\frac{1}{\sqrt{c}}$
Tương tự: $\Rightarrow VT^2\leq 3(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})$
Cũng có $\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{abc}}\leq \frac{a+b+c}{\sqrt{abc}}\leq \frac{a+b+c}{\frac{a+b+c}{3}}=3$
$\Rightarrow VT^2\leq 3.3=9\Rightarrow VT\leq 3\Rightarrow VP\geq 3\geq VT$
Dấu đẳng thức xảy ra $\Leftrightarrow a=b=c=1$