Question

Cho 2 số thực x,y thỏa mãn: \(x+y=1\). Chứng Minh:

\(x^4+y^4\ge\frac{1}{8}\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có: 

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

Lại có: \(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)

\(\Rightarrow2\left(x^2+y^2\right)\ge1\Rightarrow x^2+y^2\ge\frac{1}{2}\)

\(\Rightarrow\left(x^2+y^2\right)^2\ge\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

Vậy từ \(\left(1\right)\) có: \(2\left[\left(x^2\right)^2+\left(y^2\right)^2\right]\ge\frac{1}{4}\)

\(\Rightarrow2\left(x^4+y^4\right)\ge\frac{1}{4}\Rightarrow x^4+y^4\ge\frac{1}{8}\)

Đẳng thức xảy ra khi \(x=y=\frac{1}{2}\)