Đáp án: $ P\le 1$
Giải thích các bước giải:
Ta có:
$P=\dfrac{a}{a+\sqrt{2019a+bc}}+\dfrac{b}{b+\sqrt{2019b+ac}}+\dfrac{c}{c+\sqrt{2019c+ab}}$
$\to P=\dfrac{a}{a+\sqrt{(a+b+c)a+bc}}+\dfrac{b}{b+\sqrt{(a+b+c)b+ac}}+\dfrac{c}{c+\sqrt{(a+b+c)c+ab}}$
Vì $a+b+c=2019$
$\to P=\dfrac{a}{a+\sqrt{(a+b)(c+a)}}+\dfrac{b}{b+\sqrt{(b+a)(c+b)}}+\dfrac{c}{c+\sqrt{(c+a)(b+c)}}$
$\to P\le \dfrac{a}{a+(\sqrt{ac}+\sqrt{ba})}+\dfrac{b}{b+(\sqrt{bc}+\sqrt{ab})}+\dfrac{c}{c+(\sqrt{cb}+\sqrt{ac})}$ (bđt bunhia)
$\to P\le \dfrac{a}{a+\sqrt{ba}+\sqrt{ac}}+\dfrac{b}{\sqrt{ab}+b+\sqrt{bc}}+\dfrac{c}{\sqrt{ac}+\sqrt{cb}+c}$
$\to P\le \dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
$\to P\le \dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
$\to P\le 1$
Dấu = xảy ra khi $a=b=c=673$
