Question

Cho x, y, z >0 thỏa mãn x+y+z=1

Chứng minh : \(A=x^4+y^4+z^4\ge\dfrac{1}{27}\)

Answer 1

Với a, b, c > 0 ta có BĐT sau

\(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\) (*)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c\)

Theo BĐT (*), nếu thay \(a=x;b=y;c=z\) thì

\(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)

Theo BĐT (*), nếu thay \(a=x^2;b=y^2;c=z^2\) thì

\(x^4+y^4+z^4\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3}\ge\dfrac{\left(\dfrac{1}{3}\right)^2}{3}=\dfrac{1}{27}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=y=z\\x+y+z=1\end{matrix}\right.\) \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Answer 2

Áp dụng BĐT Cauchy Shwarz, ta có:

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

\(\Leftrightarrow x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1}{3}\)

\(\left(1+1+1\right)\left(x^4+y^4+z^4\right)\ge\left(x^2+y^2+z^2\right)^2\)

\(\Leftrightarrow x^4+y^4+z^4\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3}\ge\dfrac{1}{27}\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)