Đặt $\dfrac{1}{a}=x;\dfrac{1}{b}=y;\dfrac{1}{c}=z$
$→x^2+y^2+z^2=1$
Khi đó:
$P=\dfrac{1}{\dfrac{a(b^2+c^2)}{b^2c^2}}+\dfrac{1}{\dfrac{b(c^2+a^2)}{c^2a^2}}+\dfrac{1}{\dfrac{c(a^2+b^2)}{a^2b^2}}$
$=\dfrac{\dfrac{1}{a}}{\dfrac{1}{b^2}+\dfrac{1}{c^2}}+\dfrac{\dfrac{1}{b}}{\dfrac{1}{c^2}+\dfrac{1}{a^2}}+\dfrac{\dfrac{1}{c}}{\dfrac{1}{a^2}+\dfrac{1}{b^2}}$
$=\dfrac{x}{y^2+z^2}+\dfrac{y}{x^2+z^2}+\dfrac{z}{x^2+y^2}$
$=\dfrac{x}{1-x^2}+\dfrac{y}{1-y^2}+\dfrac{z}{1-z^2}$
Áp dụng bất đẳng thức Cauchy có: $1-x^2=\dfrac{1}{\sqrt[]2x}.\sqrt[]{2.x^2.(1-x^2)(1-x^2)}$
$\leq \dfrac{1}{\sqrt[]2x}.\sqrt[]{\dfrac{(2x^2+1-x^2+1-x^2)^3}{27}}$
$=\dfrac{1}{\sqrt[]2x}.\sqrt[]{\dfrac{8}{27}}$
$=\dfrac{2}{3.\sqrt[]3x}$
Hay $1-x^2 \leq \dfrac{2}{3\sqrt[]3x}$
suy ra $\dfrac{x}{1-x^2} \geq \dfrac{x}{ \dfrac{2}{3\sqrt[]3x}}$
$=\dfrac{3.\sqrt[]3}{2}.x^2$
Chứng minh tương tự có: $\dfrac{y}{1-y^2} \geq \dfrac{3.\sqrt[]3}{2}.y^2$
$\dfrac{z}{1-z^2} \geq \dfrac{3.\sqrt[]3}{2}.z^2$
Nên $P=\dfrac{x}{1-x^2}+\dfrac{y}{1-y^2}+\dfrac{z}{1-z^2} \geq \dfrac{3.\sqrt[]3}{2}.(x^2+y^2+z^2)=\dfrac{3.\sqrt[]3}{2}$
do $x^2+y^2+z^2=1$
Dấu $=$ xảy ra $⇔x=y=z=\dfrac{1}{\sqrt[]3}⇔a=b=c=\sqrt[]3$
Vậy $MinP=\dfrac{3.\sqrt[]3}{2}$ tại $a=b=c=\sqrt[]3$
