Đáp án + Giải thích các bước giải:
Ta có:
$\dfrac{x}{x+1}=1-\dfrac{1}{x+1}$
$\dfrac{y}{y+1}=1-\dfrac{1}{y+1}$
$\dfrac{z}{z+1}=1-\dfrac{1}{z+1}$
$⇒\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}=1-\dfrac{1}{x+1}+1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}=3-(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1})$
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta được:
$\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{z}{z+1}≥\dfrac{(1+1+1)^2}{x+y+z+3}=\dfrac{9}{4}$
$⇔-(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}≤-\dfrac{9}{4}$
$⇔3-(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1})≤\dfrac{3}{4}$
$⇔\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}≤\dfrac{3}{4}$
Dấu "=" xảy ra khi và chỉ khi $\begin{cases} x+1=y+1=z+1 \\ x+y+z=1 \end{cases}$
$⇔x=y=z=\dfrac{1}{3}$
Vậy $P_{max}=\dfrac{3}{4}⇔x=y=z=\dfrac{1}{3}$
