Question

Cho a, b  là hai số thực không âm thỏa: \(a+b\le2\)

Chứng minh \(\frac{2+a}{1+a}+\frac{1-2b}{1+2b}\ge\frac{8}{7}\)

Answer 1

VT = 2/(2+2a) + 2/(1+2b) >= 2. 4/(2+2a+1+2b) >= 8/7

 

Answer 2

\(\frac{2+a}{1+a}+\frac{1-2b}{1+2b}=\frac{\left(2+a\right)\left(1+2b\right)+\left(1-2b\right)\left(1+a\right)}{\left(1+a\right)\left(1+2b\right)}=\frac{2a+2b+3}{\left(1+a\right)\left(1+2b\right)}.\)

Ta có: \(\left(2+2a\right)\left(1+2b\right)\le\frac{\left(2+2a+1+2b\right)^2}{4}=\frac{\left(2a+2b+3\right)^2}{4}\)

\(\Rightarrow\left(1+a\right)\left(1+2b\right)\le\frac{\left(2a+2b+3\right)^2}{8}.\)

\(\Rightarrow\frac{2+a}{1+a}+\frac{1-2b}{1+2b}=\frac{2a+2b+3}{\left(1+a\right) \left(1+2b\right)}\ge\frac{2a+2b+3}{\frac{\left(2a+2b+3\right)^2}{8}}=\frac{8}{2a+2b+3}\ge\frac{8}{2.2+3}=\frac{8}{7}.\)