Question

Cho xy + yz + xz = 1. Tìm GTNN của \(x^4+y^4+z^4\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có:

\(A=x^4+y^4+z^4=\left(x^2\right)^2+\left(y^2\right)^2+\left(z^2\right)^2\)

\(\Rightarrow\left[\left(1^2\right)^2+\left(1^2\right)^2+\left(1^2\right)^2\right]\left[\left(x^2\right)^2+\left(y^2\right)^2+\left(z^2\right)^2\right]\ge\left(x^2+y^2+z^2\right)^2\)

\(\Rightarrow3\left[\left(x^2\right)^2+\left(y^2\right)^2+\left(z^2\right)^2\right]\ge\left(x^2+y^2+z^2\right)^2\)

Lại có BĐT quen thuộc \(x^2+y^2+z^2\ge xy+yz+xz\):

\(\Rightarrow3A\ge\left(x^2+y^2+z^2\right)^2\ge\left(xy+yz+xz\right)^2=1\)

\(\Rightarrow3A\ge1\Rightarrow A\ge\dfrac{1}{3}\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)