Let $a^2<2, b=2(a+1)/(a+2)$. Show $b^2<2$ (assignment) It is a part of my assignment. 
$$ \text {Let }a^2<2, \quad b=2\frac {(a+1)}{(a+2)}\quad \text{ Show } b^2<2$$ 
I already proved that a

But, I am struggling to prove $b^2<2$. 
My lecturer said that I need to manipulate $b^2$, which is larger than $b^2$ but less than 2. That is $4(a+1)^2/(a+2)^2$ < some number < 2.  
I was trying this for hours, but I couldn't find the way to solve. Thanks for helping in advance. 
 A: Hint: $\require{cancel}\;b^2 - 2 = \dfrac{4(a+1)^2}{(a+2)^2}-2=\dfrac{4a^2+\cancel{8a}+4-2a^2-\cancel{8a}-8}{(a+2)^2}=\dfrac{2a^2-4}{(a+2)^2}=\dfrac{2(a^2-2)}{(a+2)^2}\,$
A: $$\dfrac b2=1-\dfrac1{a+2}$$
$$2-\sqrt2<a+2<2+\sqrt2$$
$$1-\dfrac{2+\sqrt2}2<1-\dfrac1{a+2}<1-\dfrac{2-\sqrt2}2$$
A: same thing but with more details.
$b = \frac{(2a+1)}{a+2}$
$\frac{b}{2}  = \frac{a+1}{a+2} = \frac{a+2-1}{a+2} = = 1 - \frac{1}{a+2}$
since
$a^2 < 2$
$-\sqrt{2} < a < \sqrt{2}$
$2 -\sqrt{2} < 2 + a < 2 + \sqrt{2}$
both sides are positive so
$\frac{1}{2 + \sqrt{2}} < \frac{1}{2+a} < \frac{1}{2 - \sqrt{2}}$
$-\frac{1}{2 - \sqrt{2}} < -\frac{1}{2+a} < -\frac{1}{2 + \sqrt{2}}$
$1-\frac{1}{2 - \sqrt{2}} < 1-\frac{1}{2+a} < 1-\frac{1}{2 + \sqrt{2}}$
$1-\frac{1}{2 - \sqrt{2}} < \frac{b}{2} < 1-\frac{1}{2 + \sqrt{2}}$
$2(1-\frac{1}{2 - \sqrt{2}}) < b < 2(1-\frac{1}{2 + \sqrt{2}})$
$(1-\frac{1}{2 + \sqrt{2}})^2$
so you need to show this is less than 2 and it's QED.
A: $$a=\dfrac{2b-2}{2-b}, a^2<2\implies-\sqrt2<a<\sqrt2$$
$$\implies-\sqrt2<\dfrac{2b-2}{2-b}<\sqrt2$$
$$-\sqrt2<\dfrac{2b-2}{2-b}\iff\dfrac{2b-2+2\sqrt2-\sqrt2 b}{2-b}>0$$
$$\implies(\sqrt2b+2)(\sqrt2-1)(b-2)<0\iff-\sqrt2 <b<2\  \ \ \  (1)$$
Similarly, $$\dfrac{2b-2}{2-b}<\sqrt2\iff2<b<\sqrt2\  \ \ \  (2)$$
Combine $(1),(2)$
