Does the inequality $|\frac{1-x}{(\sqrt x +1)^2}|<1$ hold for $x>0$?

Question

I drew the graph of the function and I think the inequality is true but I don't know how prove it, can you help me please?

Answer 1

For $x<0$, the fraction is undefined.

For $x=0$ you have equality

$$\frac{1-0}{(\sqrt 0 +1)^2}=1$$

for $x >0$ you have

$$|(\sqrt x +1)^2| = x+ 2 \sqrt x +1 > x+1= |-x|+|+1|\ge|1-x|$$ (where the last inequality follows by the triangle inequality). This is equivalent to $$\left| \frac{1-x}{(\sqrt x +1)^2} \right| <1$$

Answer 2

Since there is $\sqrt{x}$ then we have $x \ge 0$.

$$|1-x| < x+1+2\sqrt{x}$$

If $x \le 1$,

$$1-x \le x+1 + 2\sqrt{x}$$

$$0 \le x + \sqrt{x}$$

If $x > 1$, we have

$$x-1 \le x+1 + 2\sqrt{x}$$

$$-1 \le \sqrt{x}$$

which is again true.

Answer 3

If $x\geq 0$, since both sides are non-negative, then the inequality is equivalent to $$(\sqrt{x}+1)^2(\sqrt{x}-1)^2=(x-1)^2\leq (\sqrt{x}+1)^4,$$ that is (note that $(\sqrt{x}+1)\not =0$), $$x-2\sqrt{x}+1=(\sqrt{x}-1)^2\leq (\sqrt{x}+1)^2=x+2\sqrt{x}+1$$ which is always satisfied for $x\geq 0$.