\(P=xz+yt+zt\\ \overset{Bunhiacopxki}{\le}\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}+zt=\dfrac{1}{\sqrt{2}}\sqrt{8\left(z^2+t^2\right)}+zt\\ \overset{Cosi}{\le}\dfrac{z^2+t^2+8}{2\sqrt{2}}+zt=\dfrac{\left(z+t\right)^2}{2\sqrt{2}}+2\sqrt{2}+\left(1-\dfrac{1}{\sqrt{2}}\right)zt\\ \overset{Cosi}{\le}4\sqrt{2}+2\sqrt{2}+\dfrac{\left(1-\dfrac{1}{\sqrt{2}}\right)\left(z+t\right)^2}{4}=4+4\sqrt{2}\)
Dấu = xảy ra khi \(x=y=\sqrt{2};z=t=2\)
Đúng 4
Bình luận (4)
