Ta có
[tex]P= \frac{3}{b+c-a}+\frac{4}{c+a-b}+\frac{5}{a+b-c} =\left ( \frac{1}{b+c-a} +\frac{1}{a+c-b}\right )+2\left ( \frac{1}{b+c-a}+\frac{1}{a+b-c} \right )+3\left ( \frac{1}{a+c-b}+\frac{1}{a+b-c} \right )[/tex]
Áp dụng BĐT [tex]\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}[/tex] có
[tex]\left ( \frac{1}{b+c-a} +\frac{1}{a+c-b}\right )+2\left ( \frac{1}{b+c-a}+\frac{1}{a+b-c} \right )+3\left ( \frac{1}{a+c-b}+\frac{1}{a+b-c} \right )\geq \frac{4}{b+c-a+a+c-b}+2.\frac{4}{b+c-a+a+b-c}+3.\frac{4}{b+a-c+a+c-b}=\frac{2}{c}+\frac{4}{b}+\frac{6}{a}=\frac{2b+4c}{bc}+\frac{6}{a}=\frac{2abc}{bc}+\frac{6}{a}=2a+\frac{6}{a}\geq 2\sqrt{2a.\frac{6}{a}}=2\sqrt{12}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
