Áp dụng bdt Bunhia ta có: $(\frac{x^{2}}{a}+\frac{y^{2}}{b}+\frac{x^{2}}{c})\geq \frac{(x+y+z)^{2}}{a+b+c}$
ta có $P=\frac{a\sqrt{ac}}{b(\sqrt{ab}+c)}+\frac{b\sqrt{ab}}{c(\sqrt{bc}+a)}+\frac{c\sqrt{bc}}{a(\sqrt{ac}+b)}$
$\Leftrightarrow$ $P=\frac{a}{b} .\frac{1}{\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}}+\frac{b}{c}.\frac{1}{\sqrt{\frac{c}{a}}+\sqrt{\frac{a}{b}}}+\frac{c}{a}.\frac{1}{\sqrt{\frac{a}{b}+\frac{b}{c}}}$
Đặt $\sqrt{\frac{a}{b}}=x $ ;$\sqrt{\frac{b}{c}}=y $ ;$\sqrt{\frac{c}{a}}=z$
$\Leftrightarrow$ $P=\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\geq$ $\frac{(x+y+z)^{2}}{2(x+y+z)}=\frac{x+y+z}{2}=\frac{1}{2}(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}})$
$\geq $ $\frac{3}{2}$$(\sqrt[3]{\sqrt{\frac{a}{b}}\sqrt{\frac{b}{c}}\sqrt{\frac{c}{a}}})=\frac{3}{2}$
dấu = xảy ra khi $x=y=z \Leftrightarrow a=b=c.$