Cho $a,b >0$ thỏa mãn : $ab+a+b=3$ tìm GTLN của biểu thức $P=\frac{3a}{b+1} +\frac{3b}{a+1} +\frac{ab}{a+b} -a^2 -b^2 $
Từ giả thiết $\Rightarrow (a+1)(b+1)=4$.
$P=\frac{3a(a+1)+3b(b+1)}{ (a+1)(b+1)} +\frac{ab}{a+b} -a^2 -b^2=\frac{3}{4}(a^2+b^2)+\frac{3}{4}(a+b)+\frac{ab}{a+b} -a^2 -b^2 $
 $P=-\frac{1}{4}(a^2+b^2)+\frac{3}{4}(a+b)+\frac{3-(a+b)}{a+b}$
 Đặt $a+b=x \implies x=3-ab \ge 3- \frac{x^2}{4}\implies x \ge 2$.
 $P=-\frac{1}{4}(x^2+2x-6)+\frac{3}{4}x+\frac{3}{x}-1=-\frac{1}{4}x^2+\frac{1}{4}x+\frac{3}{x}+\frac{1}{2}=f(x)$
 ta có $f'(x)=-\frac{2x^3-x^2+12}{4x^2} < 0    \forall x\ge 2.$
 Như vậy $f$ nghịch biến nên $f(x) \le f(2)=\frac{3}{2}$.
 Vậy $\max P =\frac{3}{2}\Leftrightarrow a=b=1.$
Hãy ấn nút tam giác màu xanh bên cạnh đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác nhé. Thanks! –  Trần Nhật Tân 25-09-12 12:30 AM
Ta có: $(a+1)(b+1)=4$
và $3\le\frac{(a+b)^2}{4}+(a+b) \Rightarrow a+b\ge 2$ .
Ta có: $P=\frac{3a(a+1)+3b(b+1)}{(a+1)(b+1)}+\frac{3}{a+b}-1-a^2-b^2$
                 $=\frac{3}{4}(a^2+b^2)+\frac{3}{4}(a+b)+ \frac{3}{a+b}-1-a^2-b^2 $
                 $= \frac{-1}{4}(a^2+b^2)+\frac{3}{4}(a+b)+ \frac{3}{a+b}-1 $ 
                 $= \frac{-1}{4}[(a+b)^2-2ab]+\frac{3}{4}(a+b)+ \frac{3}{a+b}-1 $ 
                 $= \frac{-1}{4}(a+b)^2+\frac{1}{2}[3-(a+b)]+\frac{3}{4}(a+b)+ \frac{3}{a+b}-1 $ 
                 $= \frac{-1}{4}(a+b)^2+\frac{1}{4}(a+b)+ \frac{3}{a+b}+\frac{1}{2} $
                 $=\frac{3}{2}-\frac{[(a+b)-2][(a+b)^2+(a+b)+6]}{4(a+b)}\le \frac{3}{2}$
Vậy: Max $P=\frac{3}{2} \Leftrightarrow a=b=1$ 
cả 2 cách đều rất hay. :D –  bautroitinhthuong_pt 25-09-12 11:35 PM

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