Question

Cho a,b,c >0 tm abc=1

\(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^+bc}+\frac{ac}{a^5+c^5+ac}\le1 \)

Answer 1

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

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^2+ab+b^2\right)\ge0\)*đúng*

\(\Rightarrow a^5+b^5+ab\ge a^2b^2\left(a+b\right)+ab=ab\left(ab\left(a+b\right)+1\right)\)

\(\Rightarrow\dfrac{ab}{a^5+b^5+ab}\ge\dfrac{ab}{ab\left(ab\left(a+b\right)+1\right)}=\dfrac{1}{ab\left(a+b\right)+1}\)

\(=\dfrac{c}{abc\left(a+b\right)+c}=\dfrac{c}{a+b+c}\left(abc=1\right)\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\le\dfrac{a+b+c}{a+b+c}=1=VP\)

Khi \(a=b=c=1\)