Let $a^2<2, b=2(a+1)/(a+2)$. Show $b^2<2$ (assignment)

Question

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.

Answer 1

Hint: $\require{cancel}\;b^2 - 2 = \dfrac{4(a+1)^2}{(a+2)^2}-2=\dfrac{4a^2+\cancel{8a}+4-2a^2-\cancel{8a}-8}{(a+2)^2}=\dfrac{2a^2-4}{(a+2)^2}=\dfrac{2(a^2-2)}{(a+2)^2}\,$

Answer 2

$$\dfrac b2=1-\dfrac1{a+2}$$

$$2-\sqrt2<a+2<2+\sqrt2$$

$$1-\dfrac{2+\sqrt2}2<1-\dfrac1{a+2}<1-\dfrac{2-\sqrt2}2$$

Answer 3

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.

Answer 4

$$a=\dfrac{2b-2}{2-b}, a^2<2\implies-\sqrt2<a<\sqrt2$$

$$\implies-\sqrt2<\dfrac{2b-2}{2-b}<\sqrt2$$

$$-\sqrt2<\dfrac{2b-2}{2-b}\iff\dfrac{2b-2+2\sqrt2-\sqrt2 b}{2-b}>0$$

$$\implies(\sqrt2b+2)(\sqrt2-1)(b-2)<0\iff-\sqrt2 <b<2\ \ \ \ (1)$$

Similarly, $$\dfrac{2b-2}{2-b}<\sqrt2\iff2<b<\sqrt2\ \ \ \ (2)$$

Combine $(1),(2)$