Chủ đề này có 1 trả lời

[durzaq]

x>0, y>0 thỏa mãn x+y+1=3xy.Tìm GTLN của:

$M=\frac{3x}{y(x+1)}+\frac{3y}{x(y+1)}+\frac{1}{x+y}-\frac{1}{x^2}-\frac{1}{y^2}$

[Hoang Tung 126]

Từ giả thiết :$x+y+1=3xy= > \frac{1}{y}+\frac{1}{x}+\frac{1}{xy}=3$.Đặt $\frac{1}{x}=a,\frac{1}{y}=b= > a+b+ab=3$

Ta có :$M=\frac{3x}{y(x+1)}+\frac{3y}{x(y+1)}+\frac{1}{x+y}-\frac{1}{x^2}-\frac{1}{y^2}=\frac{3b}{a+1}+\frac{3a}{b+1}+\frac{ab}{a+b}-a^2-b^2=\frac{3b(b+1)+3a(a+1)}{ab+a+b+1}+\frac{ab}{a+b}-(a^2+b^2)=\frac{3a^2+3b^2+3a+3b}{4}+\frac{ab}{a+b}-(a^2+b^2)=\frac{3a+3a-(a^2+b^2)}{4}+\frac{ab}{a+b}$(1)

Mà :$\frac{3a+3b-a^2-b^2}{4}+\frac{ab}{a+b}= \frac{-(a-1)^2-(b-1)^2+a+b+2}{4}+\frac{ab}{a+b}\leq \frac{a+b+2}{4}+\frac{ab}{2\sqrt{ab}}=\frac{a+b+2}{4}+\frac{\sqrt{ab}}{2}\leq \frac{a+b+2}{4}+\frac{\frac{ab+1}{2}}{2}=\frac{a+b+2}{4}+\frac{ab+1}{4}=\frac{a+b+ab+3}{4}=\frac{3+3}{4}=\frac{6}{4}=\frac{3}{2}$(2)

Từ (1) và (2)

$= > M\leq \frac{3}{2}= > \leq M Max=\frac{3}{2}< = > a=b=1< = > x=y=1$