[tex](a+c)(b+c)=4c^2 \Leftrightarrow (\frac{a}{c}+1)(\frac{b}{c}+1)=4[/tex]
Đặt [tex]\frac{a}{c}=x,\frac{b}{c}=y[/tex]
[tex]P=\frac{a}{b+3c}+\frac{b}{a+3c}+\frac{ab}{bc+ca}=\frac{\frac{a}{c}}{\frac{b}{c}+3}+\frac{\frac{b}{c}}{\frac{a}{c}+3}+\frac{\frac{a}{c}.\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}[/tex]
Ta thấy: [tex](x+1)(y+1)=4 \Rightarrow xy+x+y=3[/tex]
Đặt [tex]s=x+y,p=xy \Rightarrow s+p=3[/tex]
Ta có: [tex]\frac{x}{y+3}+\frac{y}{x+3}=\frac{x^2+3x+y^2+3y}{xy+3(x+y)+9}=\frac{(x+y)^2+3(x+y)-2xy}{3+2s+9}=\frac{s^2+5s-2.3}{2s+12}=\frac{s^2+5s-6}{2(s+6)}=\frac{s-1}{2}\Rightarrow P=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}=\frac{s-1}{2}+\frac{p}{s}=\frac{s-1}{2}+\frac{3-s}{s}=\frac{s}{2}+\frac{3}{s}-\frac{3}{2}\geq 2\sqrt{\frac{3}{2}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}[/tex]
Dấu "=" xảy ra khi [tex]s=\sqrt{6} \Rightarrow \left\{\begin{matrix} x+y=\sqrt{6}\\ xy=3-\sqrt{6} \end{matrix}\right.[/tex]
Lại có: [tex]s^2 \geq 4p\Rightarrow 3=s+p\leq \frac{s^2}{4}+s\Rightarrow 3> s\geq 2\Rightarrow P=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}=\frac{s-1}{2}+\frac{3-s}{s}=\frac{s^2-s+6-2s}{2s}=\frac{s^2-3s+6}{2s}=\frac{(s-2)(s-3)}{2s}+1\leq 1[/tex]
Dấu "=" xảy ra khi [tex]s=2 \Rightarrow x=y=1 \Rightarrow a=b=c[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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