Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$

Question

Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$

The book I am using asserts a non-trivial way of proving this inequality, but I cannot see why this cannot be proven by rearranging the statement equally as rigorously.

Let $$\frac{2k+2}{2k+3}=\frac{2k}{2k+1}, k\in\mathbb N$$ $$(2k+2)(2k+1)=2k(2k+3)$$ $$2k^2+3k+1=2k^2+3k$$ $$1=0$$ Contradiction, therefore no solution to where the two sides are equal

Let k=1 $$\frac{2+2}{2+3}= 0.8$$ $$\frac{2}{3}=0.66666$$ Since there are no points of intersection, and for k=1 LHS>RHS, and both sides are continuous on k>0, inequality must hold.

Answer 1

$\frac{2x+2}{2x+3}>\frac{2x}{2x+1}, X\in\mathbb R^+$

$ f: x \rightarrow \frac{2x}{2x+1} = f: x \rightarrow 1- \frac{1}{2x+1}$ is easily demonstrable to be strictly increasing

$x > y \rightarrow 2x+1 > 2y+1 \rightarrow \frac{1}{2x+1} <\frac{1}{2y+1} \rightarrow 1- \frac{1}{2x+1} > 1-\frac{1}{2y+1}$ therefore $f(k+1) > f(k) $

i.e : $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N \subset \mathbb R^+$

Answer 2

Let's consider the following inequality verified for every k natural: $$\frac{2}{(2k+3)(2k+1)}\gt0$$ Now, with simple manipulations $$\frac{4k^2+6k+2-4k^2-6k}{(2k+3)(2k+1)}\gt0$$ $$\frac{(2k+2)(2k+1)-2k(2k+3)}{(2k+3)(2k+1)}\gt0$$ $$\frac{2k+2}{2k+3}-\frac{2k}{2k+1}\gt0$$ $$\frac{2k+2}{2k+3}\gt\frac{2k}{2k+1}$$

Answer 3

Because $k\in N$, and the denominators are always positive, we have: $$(2k+2)(2k+1)>2k(2k+3)$$ So: $$4k^2+6k+2>4k^2+6k$$ Now, we can cancel out $4k^2$ and $6k$, obtaining:$$2>0$$ which is always true. The inequality holds.