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.
 A: Your process is correct until the 4th step. There is a sign error in the 5th step. Gigili's answer provides the correct solution, but if you are still stumped about where the minus sign is coming from, here is a more detailed look at all the manipulations and properties involved to get from the 4th step in your process to the correct 5th step (which is the 4th step in Gigili's answer).
$$\eqalignno{
&\phantom{=}\thinspace 5\left(1 -{k^2 + k + 2\over k{(k + 1)}^2} \right)&(1)\cr
&=5\left(1 + \left(-{k^2 + k + 2\over k{(k + 1)}^2}\right) \right)&(2)\cr
&=5\left(1 + \left(-1\left({k^2 + k + 2\over k{(k + 1)}^2}\right)\right) \right)&(3)\cr
&=5\left(1 + \left(-1\left({k(k + 1) + 2\over k{(k + 1)}^2}\right)\right) \right)&(4)\cr
&=5\left(1 + \left(-1\left({k(k + 1)\over k{(k + 1)}^2} + {2\over k{(k + 1)}^2}\right)\right) \right)&(5)\cr
&=5\left(1 + \left(\biggl(-1\biggr)\left({k(k + 1)\over k{(k + 1)}^2}\right) + \biggl(-1\biggr)\left( {2\over k{(k + 1)}^2}\right)\right) \right)&(6)\cr
&=5\left(1 + \left(\left(-{k(k + 1)\over k{(k + 1)}^2}\right) + \left(- {2\over k{(k + 1)}^2}\right)\right)\right)&(7)\cr
&=5\left(1 + \left(-{k(k + 1)\over k{(k + 1)}^2}\right) + \left(- {2\over k{(k + 1)}^2}\right)\right)&(8)\cr
&=5\left(1 - {k(k + 1)\over k{(k + 1)}^2} - {2\over k{(k + 1)}^2} \right)&(9)\cr
}$$
(1) Given (4th step in the incorrect solution / 3rd step in Gigili's answer).
(2) Definition of subtraction in terms of addition.
(3) Multiplication property of negative one.
(4) Associative property of addition; factoring.
(5) Addition of fractions with like denominators.
(6) Distributive property.
(7) Multiplication property of negative one.
(8) Associative property of addition.
(9) Definition of subtraction in terms of addition (similar to 5th step in the incorrect solution, but with correct sign / 4th step in Gigili's answer).
A: $$=5(1+\frac1{(k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$=5(1+\frac{k}{k(k+1)^2}-\frac{(k+1)^2}{k(k+1)^2}-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}\color{red}{-}\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}\color{red}{-}\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}\color{red}{-10}\frac1{k(k+1)^2}$$

Where $k \geq 2$ , so $$\frac{-10}{k(k+1)^2} < 0$$
Therefore
$$5 - \frac5{k+1}-\frac{10}{k(k+1)^2} \leq 5 - \frac5{k+1}$$

$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$ Which is the desired result.
A: 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 $k \geq 2$.
