Áp dụng BĐT bunyakovsky:
\(\left(x^2+y^2+2z^2+2t^2\right)\left(1+1+\dfrac{1}{2}+\dfrac{1}{2}\right)\ge\left(x+y+z+t\right)^2\)
Lại có: theo AM-GM:\(\left(x+y+z+t\right)^2\ge4\left(x+z\right)\left(y+t\right)\)
\(\Rightarrow4VT\le3\Leftrightarrow VT\le\dfrac{3}{4}\)
Dấu = xảy ra khi \(x=y=2z=2t=\dfrac{1}{\sqrt{3}}\)
P/s : Nếu đề mà cho là (x+y)(z+t) thì die :v