\(\sqrt{2019a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ab}+\sqrt{ac}\left(bunhia\right)\)
\(\Rightarrow\dfrac{a}{a+\sqrt{2019a+bc}}\le\dfrac{a}{a+\sqrt{ab}+\sqrt{ac}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
\(tương\) \(tự\Rightarrow P\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\) \(dấu"="\Leftrightarrow a=b=c=673\)
\(1\) \(cách\) \(khác\) \(\sqrt{2019a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{a+b+a+c}{2}=\dfrac{2a+b+c}{2}\)
\(\dfrac{a}{a+\sqrt{2019a+bc}}=\dfrac{a\left(a-\sqrt{2019a+bc}\right)}{a^2-\left(a+b+c\right)a-bc}=\dfrac{a\left(a-\sqrt{2019a+bc}\right)}{-\left(ab+ca+bc\right)}=\dfrac{a\sqrt{\left(a+b\right)\left(b+c\right)}-a^2}{ab+bc+ca}\le\dfrac{a\left(\dfrac{2a+b+c}{2}\right)-a^2}{ab+bc+ca}=\dfrac{\dfrac{2a^2+ab+ac}{2}-a^2}{ab+bc+ca}=\dfrac{\dfrac{ab+ac}{2}}{ab+bc+ca}\)
\(tương\) \(tự:\dfrac{b}{b+\sqrt{2019b+ca}}\le\dfrac{\dfrac{ba+bc}{2}}{ab+bc+ca};\dfrac{c}{c+\sqrt{2019c+ab}}\le\dfrac{\dfrac{ca+cb}{2}}{ab+bc+ca}\Rightarrow P\le\dfrac{1}{2}\dfrac{2\left(ab+bc+ca\right)}{ab+bc+ca}=1\)
