Dễ thấy c khác 0. Ta có: [tex](a+c)(b+c)=4c^2\Rightarrow (\frac{a}{c}+1)(\frac{b}{c}+1)=4[/tex]
Đặt [tex]\frac{a}{c}=x,\frac{b}{c}=y\Rightarrow (x+1)(y+1)=4\Rightarrow xy+x+y=3[/tex]
Ta thấy: [tex]P=\frac{a}{b+3c}+\frac{b}{a+3c}+\frac{ab}{bc+ac}=\frac{\frac{a}{c}}{\frac{b}{c}+3}+\frac{\frac{b}{c}}{\frac{a}{c}+3}+\frac{\frac{a}{c}.\frac{b}{c}}{\frac{a}{c}+\frac{b}{c}}=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}=\frac{x^2+3x+y^2+3y}{xy+3(x+y)+9}+\frac{xy}{x+y}[/tex]
Đặt [tex]t=x+y\Rightarrow xy=3-t[/tex]. Lại có: [tex](x+y)^2\geq 4xy\Rightarrow t^2\geq 4(3-t)\Rightarrow t^2+4t-12\geq 0\Rightarrow (t-2)(t+6)\geq 0\Rightarrow t\geq 2[/tex]
[tex]P=\frac{x^2+3x+y^2+3y}{xy+3(x+y)+9}+\frac{xy}{x+y}=\frac{(x+y)^2-2xy+3t}{(3-t)+3t+9}+\frac{3-t}{t}=\frac{t^2-2(3-t)+3t}{2t+12}+\frac{3-t}{t}=\frac{t^2+5t-6}{2t+12}+\frac{3-t}{t}=\frac{(t-1)(t+6)}{2(t+6)}+\frac{3-t}{t}=\frac{t-1}{2}+\frac{3-t}{t}=\frac{t}{2}+\frac{3}{t}-\frac{1}{2}-1\geq 2\sqrt{\frac{t}{2}.\frac{3}{t}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}[/tex]
Dấu "=" xảy ra khi [tex]\frac{t}{2}=\frac{3}{t}\Leftrightarrow t=\sqrt{6}\Leftrightarrow \left\{\begin{matrix} x+y=\sqrt{6}\\ xy=3-\sqrt{6} \end{matrix}\right.[/tex]
Lại có: [tex]P=\frac{t-1}{2}+\frac{3-t}{t}-\frac{t^2-t+6-2t}{2t}=\frac{t^2-3t+6}{2t}=\frac{(t-2)(t-3)+2t}{2t}=\frac{(t-2)(t-3)}{2t}+1[/tex]
Vì [tex]xy=3-t> 0\Rightarrow t< 3\Rightarrow (t-2)(t-3)\leq 0\Rightarrow P\leq 1[/tex]
Dấu "=" xảy ra khi [tex]t=2\Leftrightarrow x=y=1[/tex]
Vậy Min P = [tex]\sqrt{6}-\frac{3}{2}[/tex], Max P = 1.
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