Đổi biến $(x;y;z;t)→(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c};\dfrac{1}{d})$
$→abcd=1$
Khi đó $VT=∑\dfrac{xyzt}{x^3(yz+zt+ty)}=∑\dfrac{\dfrac{1}{x^3}}{\dfrac{1}{xt}+\dfrac{1}{xy}+\dfrac{1}{xz}}=∑\dfrac{a^3}{ad+ab+ac}=∑\dfrac{a^2}{b+c+d}≥\dfrac{(a+b+c+d)^2}{3(a+b+c+d)}=\dfrac{a+b+c+d}{3}≥\dfrac{4\sqrt[4]{abcd}}{3}=\dfrac{4}{3}$
Dấu $=$ xảy ra $⇔a=b=c=d$