Giải thích các bước giải:
$P=(a^2+b^2+c^2)(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})$
$\rightarrow P\ge (\dfrac{(a+b)^2}{2}+c^2)(\dfrac{1}{2}(\dfrac{1}{a}+\dfrac{1}{b})^2+\dfrac{1}{c^2})$
$\rightarrow P\ge (\dfrac{(a+b)^2}{2}+c^2)(\dfrac{1}{2}(\dfrac{4}{a+b})^2+\dfrac{1}{c^2})$
$\rightarrow P\ge (\dfrac{(a+b)^2}{2}+c^2)(\dfrac{8}{(a+b)^2}+\dfrac{1}{c^2})$
Đặt $a+b=x,c=y\rightarrow x\le y$
$\rightarrow P\ge (\dfrac{x^2}{2}+y^2)(\dfrac{8}{x^2}+\dfrac{1}{y^2})$
$\rightarrow P\ge 4+1+\dfrac{8y^2}{x^2}+\dfrac{x^2}{2y^2}$
$\rightarrow P\ge 5+\dfrac{15y^2}{2x^2}+\dfrac{1}{2}.(\dfrac{y^2}{x^2}+\dfrac{x^2}{y^2})$
$\rightarrow P\ge 5+\dfrac{15}{2}+\dfrac{1}{2}.2\sqrt{\dfrac{y^2}{x^2}.\dfrac{x^2}{y^2}}(y\ge x)$
$\rightarrow P\ge 5+\dfrac{15}{2}+\dfrac{1}{2}.2$
$\rightarrow P\ge \dfrac{27}{2}$
Dấu = xảy ra khi $a=b=\dfrac{c}{2}$
