Với x;y dương, ta có BĐT:
\(x^5+y^5\ge x^2y^2\left(x+y\right)\)
Thật vậy, BĐT tương đương:
\(x^5-x^4y+y^5-xy^4\ge0\)
\(\Leftrightarrow x^4\left(x-y\right)-y^4\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (luôn đúng)
Áp dụng:
\(\Rightarrow A\le\dfrac{ab}{a^2b^2\left(a+b\right)+ab}+\dfrac{bc}{b^2c^2\left(b+c\right)+bc}+\dfrac{ca}{c^2a^2\left(c+a\right)+ca}\)
\(A\le\dfrac{1}{ab\left(a+b\right)+1}+\dfrac{1}{bc\left(b+c\right)+1}+\dfrac{1}{ca\left(c+a\right)+1}\)
\(A\le\dfrac{abc}{ab\left(a+b\right)+abc}+\dfrac{abc}{bc\left(b+c\right)+abc}+\dfrac{abc}{ca\left(c+a\right)+abc}=\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}=1\)
