Prove that for all nonnegative real numbers $x, \frac {2|x-3|}{x+1} \le 7$

Question

The question also says to consider two cases, based on the definition of absolute value. I am not sure how to prove this problem.

Answer 1

Prove that for all nonnegative x: $\frac{2|x-3|}{x+1} \leq 7$

We need to consider two cases:

Case 1: $x-3\geq 0$

$\implies x\geq 3, |x-3|=x-3$

Then:

$\frac{2|x-3|}{x+1} = \frac{2x-6}{x+1}\leq 7 \implies 2x-6 \leq 7x+7 \implies 5x\geq -13$

Clearly, since x is bigger than three, the latter equation is satisfied, and so the proof is done for case 1.

Case 2: $x-3<0$ $ \implies x<3, |x-3|=3-x$

Then

$\frac{2|x-3|}{x+1} = \frac{6-2x}{x+1}\leq 7 \implies 6-2x\leq 7x+7 \implies 9x\geq -1 \implies x\geq -1/9$

So to satisfy the condition in case $2$, $x$ has to be larger than $-1/9$. Indeed, by assumption, $x\geq 0$, so that condition is easily satisfied.

This finishes the proof since there are no further cases (the absolute value function has only 2 cases)