Cách 3:
Đặt $a=\frac{yzt}{x^3};y=...;z=...;t=...$
BĐT cần c/m trở thành:
$\Sigma \frac{x^6}{(x^3+yzt)^2}\geq 1$
A/d Cauchy-Schwarz:
$\frac{x^6}{(x^3+yzt)^2}\geq \frac{(x^3+y^3+..)}{\Sigma (x^3+yzt)^3}$
Ta cần c/m:
$(x^3+y^3+z^3+t^3)^2\geq \Sigma (x^3+yzt)^2$
hay $2\Sigma x^3y^3\geq 2\Sigma x^3yzt+\Sigma y^2z^2t^2$
mà theo AM-GM:
$2\Sigma x^3yzt\leq 2/3.\Sigma x^3(y^3+z^3+t^3)=4/3.\Sigma x^3+y^3$
=>.....