Suppose $0\leq{t}\leq{x}\leq1$ and let $z=\frac{t}{x}$ prove $|1-z|\leq1$

Question

Suppose $0\leq{t}\leq{x}\leq1$ and let $z=\frac{t}{x}$ prove $|1-z|\leq1$

I am hoping someone can help me with this question I'm not sure what technique is used to prove an inequality holds. I have only proven inequalities by induction which doesn't seem to be the case for this inequality, and my textbook doesn't have any examples like this.

I figure it's just some algebraic manipulations which need to be performed but I don't know what I'm supposed to get. If the question had asked to disprove an inequality I know I would just have to find a contradiction but since it says to prove it I'm not sure what to do.

I am also asked to prove $$\frac{1}{|1+xz|}\leq1$$$$|\frac{x-xz}{1+xz}|^n\leq|x|^n$$ for $n\in\mathbb{N}$

and also $$|1+t|^{-\frac{1}{2}}\leq1$$

So this is what I tried for the first inequality:

$$|1-z|\leq1$$

$$=1-\frac{t}{x}\leq1$$

It seems pretty obvious that if I subtract a positive fraction from 1 it result in a value less than 1 but is that really a proof?

For the second inequality $$\frac{1}{|1+xz|}\leq1$$$$=\frac{1}{1+t}\leq1$$

Again this one seems obvious that 1 divided by a value greater than 1 is less than 1 but it doesn't sound like a formal way of saying it.

Any help would be greatly appreciated.

Thank you,

Answer 1

Just use definition of $|A| \le B \iff - B\le A \le B$ with $A = 1- z, B = 1$. Thus you check: $-1 \le 1-z \le 1\iff 0 \le z \le 2$, which is true since $z \ge 0$, and $z \le 1 < 2 \implies z < 2$.