Đáp án:
$\max Q = 1 \Leftrightarrow x = y = z = \dfrac{2020}{3}$
Giải thích các bước giải:
$\begin{array}{l}\text{Ta có:}\\ \quad \dfrac{x}{x + \sqrt{2020x + yz}}\\ =\dfrac{x}{x + \sqrt{(x+y+z)x + yz}}\\ = \dfrac{x}{x+\sqrt{x^2 + xy + zx + yz}}\\ = \dfrac{x}{x+ \sqrt{(x+y)(x+z)}}\\ Do\,\,\sqrt{(x+y)(x+z)} \geq \sqrt{xy} + \sqrt{xz}\qquad (Bunyakovsky)\\ nên\,\,\dfrac{x}{x+\sqrt{(x+y)(x+z)}} \leq \dfrac{x}{x+\sqrt{xy} + \sqrt{xz}}\\ hay\,\,\dfrac{x}{x + \sqrt{2020x + yz}} \leq \dfrac{\sqrt x}{\sqrt x + \sqrt y + \sqrt z}\\ \text{Hoàn toàn tương tự, ta được:}\\ \dfrac{y}{y+ \sqrt{2020y + zx}} \leq \dfrac{\sqrt y}{\sqrt y + \sqrt z + \sqrt x}\\ \dfrac{z}{z+\sqrt{2020z + xy}}\leq \dfrac{\sqrt z}{\sqrt z + \sqrt x + \sqrt y}\\ \text{Cộng vế theo vế ta được:}\\ \dfrac{x}{x + \sqrt{2020x + yz}}+\dfrac{y}{y+ \sqrt{2020y + zx}} +\dfrac{z}{z+\sqrt{2020z + xy}}\leq \dfrac{\sqrt x + \sqrt y + \sqrt z}{\sqrt x + \sqrt y+ \sqrt z} = 1\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x = y= z = \dfrac{2020}{3}\\ Vậy\,\,\max Q = 1 \Leftrightarrow x = y = z = \dfrac{2020}{3} \end{array}$
