Proving $\frac{1}{\sqrt{x}}\ge \frac{2}{x+1}$ for $x> 0$

Question

Prove:

$$\frac{1}{\sqrt{x}}\ge \frac{2}{x+1}, \quad\forall x>0$$

Yeah, pretty much it. I've tried all manner of rearranging and just can't seem to get it. Thanks.

Answer 1

Hint: if $a,b\geq 0$ then you can use $$a+b\geq 2\sqrt{ab}$$

and notice that if $a\ne b$, then we have $>$ instead of $\geq$

Answer 2

Substitute $y:=\sqrt x>0$, multiply away the (positive!) denominators, bring all terms to the left side and recognize the left as a square.

Answer 3

It is often a good idea to just multiply with all the denominators, and see if you can do something with the more complicated expressions. Try such trivial approaches first, and if you get stuck, you can still ask for help.

Answer 4

by cross multiplying, we get the following question:

$$ x+1 \geq 2\sqrt{x} $$

or

$$x - 2\sqrt{x} + 1 \geq 0 $$

Now we can say:

$$x-2\sqrt{x}+1 = \left(\sqrt{x} - 1\right)^2 \geq 0 $$

which proves the inequality :)