Question

Cho a,b\(\ge\)0 thỏa mãn a+b\(\le\)2.

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

Answer 1

cho hỏi viết phân số kiểu gị

Answer 2

$\frac{2+a}{1+a}=1+\frac{1}{1+a}$
\(\frac{1-2b}{1+2b}=-1+\frac{2}{1+2b}\)$\frac{1-2b}{1+2b}=-1+\frac{2}{1+2b}$
$\frac{1}{1+a}+\frac{2}{2+2b}=\frac{2}{2+2a}+\frac{2}{2+2b}\ge \frac{8}{4+2\left(a+b\right)}=\frac{8}{7}$

Answer 3

$\frac{1}{1+a}+\frac{2}{2+2b}=\frac{2}{2+2a}+\frac{2}{2+2b}\ge \frac{8}{4+2\left(a+b\right)}=\frac{8}{7}$

Answer 4

\(\frac{2+a}{1+a}=1+\frac{1}{1+a}\)

Answer 5

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

Answer 6

\(VT=\frac{2}{2+2a}+\frac{2}{1+2b}\ge2.\frac{4}{2+2a+1+2b}\ge\frac{8}{7}\) (đpcm)