Đáp án:
$\max\left(\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\right) = \dfrac{3}{4} \Leftrightarrow x = y = z = \dfrac{1}{3}$
Giải thích các bước giải:
Ta có:
$\dfrac{x}{x+1} = 1 - \dfrac{1}{x+1}$
$\dfrac{y}{y+1} = 1 - \dfrac{1}{y+1}$
$\dfrac{z}{z+1} = 1 - \dfrac{1}{z+1}$
Do đó:
$\dfrac{x}{x+1} +\dfrac{y}{y+1} +\dfrac{z}{z+1} = 3 - \left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)$
Áp dụng bất đẳng thức $Cauchy-Schwarz$ dạng $Engel$ ta được:
$\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1} \geq \dfrac{(1+1+1)^2}{x+y+z+3} = \dfrac{9}{4}$
$\Leftrightarrow - \left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right) \leq - \dfrac{9}{4}$
$\Leftrightarrow 3 - \left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right) \leq \dfrac{3}{4}$
$\Leftrightarrow \dfrac{x}{x+1} +\dfrac{y}{y+1} +\dfrac{z}{z+1} \leq \dfrac{3}{4}$
Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}x+1 = y + 1 = z + 1\\x + y + z = 1\end{cases}\Leftrightarrow x = y = z = \dfrac{1}{3}$
Vậy $\max\left(\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\right) = \dfrac{3}{4} \Leftrightarrow x = y = z = \dfrac{1}{3}$
