[tex]\frac{1}{2}.\sqrt{\frac{a}{a+1}}\leq \frac{1}{2}(\frac{a}{a+1}+\frac{1}{4})\Rightarrow \sqrt{\frac{a}{a+1}}\leq \frac{a}{a+1}+\frac{1}{4}[/tex]
Tương tự, cộng vế theo vế ta có: [tex]P\leq (\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1})+\frac{3}{4}[/tex]
Lại có: [tex]\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq \frac{9}{a+b+c+3}=\frac{9}{4}\Rightarrow \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\leq 3-\frac{9}{4}=\frac{3}{4}\Rightarrow P\leq \frac{3}{2}[/tex]
Dấu "=" xảy ra tại [tex](a,b,c)=(\frac{1}{3},\frac{1}{3},\frac{1}{3})[/tex]
Ta thấy: [tex]a \leq 1\Rightarrow \frac{a}{a+1}\leq \frac{1}{2}\Rightarrow \sqrt{\frac{a}{a+1}}\geq \frac{\sqrt{2}a}{a+1}[/tex]
Tương tự cộng vế theo vế ta có: [tex]P \geq \sqrt{2}(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1})[/tex]
Mà [tex]3-(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1})=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)}{(a+1)(b+1)(c+1)}=\frac{ab+bc+ca+2(a+b+c)+3}{abc+ab+bc+ca+a+b+c+1}=\frac{ab+bc+ca+5}{abc+ab+bc+ca+2}\leq \frac{ab+bc+ca+5}{ab+bc+ca+2}=1+\frac{3}{ab+bc+ca+2} \leq 1+\frac{3}{2}=\frac{5}{2}\Rightarrow \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3-\frac{5}{2}=\frac{1}{2}\Rightarrow P\geq \frac{\sqrt{2}}{2}[/tex]
Dấu "=" xảy ra tại [tex](a,b,c)=(1,0,0)[/tex] và các hoán vị.
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
