Đáp án: $A\le 1$
Giải thích các bước giải:
Ta có:
$a^5+a^5+b^5+b^5+b^5\ge 5\sqrt{a^5\cdot a^5\cdot b^5\cdot b^5\cdot b^5}=5a^2b^3$
$a^5+a^5+a^5+b^5+b^5\ge 5\sqrt{a^5\cdot a^5\cdot a^5\cdot b^5\cdot b^5}=5a^3b^2$
Cộng vế với vế
$\to 5(a^5+b^5)\ge 5a^2b^3+5a^3b^2$
$\to a^5+b^5\ge a^2b^2(a+b)$
Tương tự suy ra :
$b^5+c^5\ge b^2c^2(b+c)$
$c^5+a^5\ge c^2a^2(c+a)$
$\to A=\dfrac{ab}{a^5+b^5+ab}+\dfrac{bc}{b^5+c^5+bc}+\dfrac{ca}{c^5+a^5+ca}$
$\to A\le \dfrac{ab}{a^2b^2(a+b)+ab}+\dfrac{bc}{ b^2c^2(b+c)+bc}+\dfrac{ca}{c^2a^2(c+a)+ca}$
$\to A\le \dfrac{1}{ab(a+b)+1}+\dfrac{1}{ bc(b+c)+1}+\dfrac{1}{ ca(c+a)+1}$
$\to A\le \dfrac{1}{ab(a+b)+abc}+\dfrac{1}{ bc(b+c)+abc}+\dfrac{1}{ ca(c+a)+abc}$
$\to A\le \dfrac{1}{ab(a+b+c)}+\dfrac{1}{ bc(a+b+c)}+\dfrac{1}{ ca(c+a+b)}$
$\to A\le \dfrac{c}{abc(a+b+c)}+\dfrac{a}{ abc(a+b+c)}+\dfrac{b}{ cab(c+a+b)}$
$\to A\le \dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{ a+b+c}$
$\to A\le 1$
