Question

[Roxie]

chứng minh rằng \(\frac{n+1}{n}>\frac{n+3}{n+1}\)

Answer 1

[Akai Haruma]

Lời giải:

ĐK: $0< n< 1$

Ta thấy:

$\frac{n+1}{n}-\frac{n+3}{n+1}=\frac{(n+1)(n+1)-n(n+3)}{n(n+1)}$

$=\frac{n^2+2n+1-n^2-3n}{n(n+1)}=\frac{1-n}{n(n+1)}>0$ với mọi $0< n< 1$

Do đó $\frac{n+1}{n}>\frac{n+3}{n+1}$ (đpcm)