Proving $(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$ Can anyone tell me what I am doing wrong? need to prove for $k\ge2$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$$$(5-\frac5k )(1+\frac{1}{(k+1)^2})= 5(1-\frac1k)(1+\frac1{(k+1)^2})$$
$$=5(1+\frac1{k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}+\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}+\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}+\frac{10}{k(k+1)^2}\le5-\frac5{k+1}$$
which doesn't look true.

Let $f(x) = (5-\frac{5}{x})(1+\frac{1}{(1+x)^2})-5+\frac{5}{1+x}$. After a little algebra, this gives $f(x) = -\frac{10}{x(x+1)^2}$. Hence $f(x) \leq  0$ when $x>0$.
Hence $f(k) \leq 0$ for $k \geq 2$.
