Prove $|1+t|^{-\frac{1}{2}}\le1$

Question

I am having a hard time proving this inequality hopefully someone here may be able to help me.

$0<t\leq{x}<1$ and let $z=\frac{t}{x}$

Prove $$|1+t|^{-\frac{1}{2}}\le1$$$$\implies|1+{z}{x}|^{-\frac{1}{2}}\le1$$$$\implies\frac{1}{|\sqrt{1+{z}{x}}|}\le1$$$$\implies1\le|\sqrt{1+{z}{x}|}$$

Not really sure where to go from here. I can do all sorts of algebraic manipulations I just have no idea when one of these inequality proofs is finished.

I don't think I'm allowed to just say it's obvious that $$1\le|\sqrt{1+{z}{x}|}$$.

Any advice regarding inequality proofs would be greatly appreciated.

Thank you,

Answer 1

From your working, it seems that you meant to say $t=\frac{z}x$ rather than $z=\frac{t}{x}$.

If $t \ge 0$, then we have $$1 \le 1+t$$

Square root is an increasing fucntion.

$$\sqrt{1} \le \sqrt{1+t}$$

$$1 \le \sqrt{1+t}$$

Answer 2

Jack, actually we can use proof without $x$.

If $0 \le t <1$ then $$ 1+t \ge 1 \implies \sqrt{|1+t|} \ge 1 \implies \frac{1}{ \sqrt{|1+t|} } \le 1 $$

So even if we have statement $ 0 \le t \le x < 1$, the above still applies.