Áp dụng bđt AM-GM ta có
$T=zt(y+s-x)+x(s+z)(y+t)\leq \frac{(z+t+y+s-x)^3}{27}+x\frac{(y+z+t+s)^2}{4}=(\frac{1-2x}{3})^3+x(\frac{1-x}{2})^2\leq \frac{1}{5} $(do $\frac{1}{5}-[\left ( \frac{1-2x}{3} \right )^3+x\left ( \frac{1-x}{2} \right )^2]=\frac{\left ( 5x-1 \right )^2\left (8+5 x \right )}{2700}\geq 0$)
Dấu "=" xảy ra khi $x=y=z=t=s=\frac{1}{5}$