Đáp án:
Pmin=4
Giải thích các bước giải:
\(\begin{array}{l}
\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\\
\Rightarrow \frac{2}{b} = \frac{{a + c}}{{ac}}\\
\Rightarrow 2ac = ab + bc\\
P = \frac{{a + b}}{{2a - b}} + \frac{{c + b}}{{2c - b}}\\
P = \frac{{ac + bc}}{{2ac - bc}} + \frac{{ac + ab}}{{2ac - ab}}\\
P = \frac{{ac + bc}}{{ab}} + \frac{{ac + ab}}{{bc}}\\
P = \frac{1}{2}.(\frac{{2ac + 2bc}}{{ab}} + \frac{{2ac + 2ab}}{{bc}})\\
P = \frac{1}{2}.(\frac{{ab + 3bc}}{{ab}} + \frac{{bc + 3ab}}{{bc}})\\
P = \frac{1}{2}.(2 + 3.(\frac{c}{a} + \frac{a}{c}))
\end{array}\)
Áp dụng BĐT Cosi cho 2 số dương ta có
\(\begin{array}{l}
\frac{a}{c} + \frac{c}{a} \ge 2\sqrt {\frac{a}{c}.\frac{c}{a}} = 2\\
\Rightarrow P \ge 4
\end{array}\)
Dấu "=" xảy ra ⇔a=b=c
