Question

cho x, y, z>= √2. chứng minh rằng 1/y*5+z*5+2xyz + 1/z*5+x*5+2xyz + 1/x*5+y*5+2xyz <= 1/2xyz

Answer 1

Ta có: \(a^5+b^5\ge a^2b^2\left(a+b\right)\)

\(\Leftrightarrow a^5+b^5+2abc\ge a^2b^2\left(a+b\right)+2abc\)

\(\ge ab\left[ab\left(a+b\right)+2c\right]\ge ab\left[2\left(a+b\right)+2c\right]=2ab\left(a+b+c\right)\) (áp dụng với \(a,b,c\ge\sqrt{2}\))

\(\Rightarrow\frac{1}{a^5+b^5+2abc}\le\frac{1}{2ab\left(a+b+c\right)}\)

Áp dụng vào bài toán ta được

\(P\le\frac{1}{2xy\left(x+y+z\right)}+\frac{1}{2yz\left(x+y+z\right)}+\frac{1}{2zx\left(x+y+z\right)}\)

\(=\frac{x+y+z}{2xyz\left(x+y+z\right)}=\frac{1}{2xyz}\)