Question

Cho a, b , c là 3 số thực dương thỏa mãn abc=1. Tính GTLN của

P= \(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\)

Answer 1

Với các số dương x; y ta có:

\(x^5+y^5=\left(x^3+y^3\right)\left(x^2+y^2\right)-x^2y^2\left(x+y\right)\)

\(\Rightarrow x^5+y^5\ge xy\left(x+y\right).2xy-x^2y^2\left(x+y\right)=x^2y^2\left(x+y\right)\)

\(\Rightarrow P\le\frac{ab}{a^2b^2\left(a+b\right)+ab}+\frac{bc}{b^2c^2\left(b+c\right)+bc}+\frac{ca}{c^2a^2\left(c+a\right)+ca}\)

\(P\le\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}+\frac{1}{ca\left(c+a\right)+1}\)

\(P\le\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}+\frac{abc}{ca\left(a+c\right)+abc}\)

\(P\le\frac{c}{a+b+c}+\frac{a}{a+b+c}+\frac{b}{a+b+c}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)