Đáp án: $A\le 1$
Giải thích các bước giải:
Ta có:
$\sqrt{2021a+bc}=\sqrt{(a+b+c)a+bc}$ vì $a+b+c=2021$
$\to\sqrt{2021a+bc}=\sqrt{(a+b)(c+a)}\ge \sqrt{(\sqrt{ac}+\sqrt{ba})^2}=(\sqrt{ac}+\sqrt{ba})$
$\to a+\sqrt{2021a+bc}\ge a+\sqrt{ac}+\sqrt{ba}$
$\to a+\sqrt{2021a+bc}\ge \sqrt{a}(\sqrt{a}+\sqrt{b}+\sqrt{c})$
$\to \dfrac{a}{ a+\sqrt{2021a+bc}}\le \dfrac{a}{ \sqrt{a}(\sqrt{a}+\sqrt{b}+\sqrt{c})}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
Tương tự
$\dfrac{b}{b+\sqrt{2021b+ac}}\le \dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
$\dfrac{c}{c+\sqrt{2021c+ab}}\le \dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
Cộng vế với vế
$\to A\le \dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
$\to A\le 1$
Dấu = xảy ra khi $a=b=c=\dfrac{2021}{3}$
