Proofs involving strict inequalities

Question

let $a,b \in R$

Prove that if $3 \lt a \lt 5$ and $b= 2 + \sqrt{a-2}$ then, $3 \lt b \lt a$

My approach was simply to start with the first inequality and transform it into b and see what happens.

Subtracting 2 gives: $1 \lt a-2 \lt 3$

Taking the square root gives: $1 \lt \sqrt{a-2} \lt \sqrt{3}$

Adding two gives: $3 \lt 2 + \sqrt{a-2} \lt \sqrt{3}+2$

thus $3 \lt b \lt \sqrt{3}+2$

So the interval on b is in fact smaller than the interval on a but that just doesn't seem like it is enough. That's not a very convincing argument. Isn't it plausible that b could equal 3.11 and a could equal 3.10 or something like that. Those numbers fall in the inside the allowed intervals.

Plus this method didn't work on the other 4 questions in the problem.

Suppose $ a \gt 2$ and $b = 1 + \sqrt{a-1}$ using the same technique lands

$b$ as $b \gt 2$

Answer 1

You are asked to prove that if $3<a<5$ and $b=2+\sqrt{a-2}$ then $3<b<a$.

You proved $3<b$ (and $b<\sqrt3+2$), but you didn't prove $b<a$.

$b<a$ means $\sqrt{a-2}+2<a$ or $\sqrt{a-2}<a-2$.

This suggests letting $x=a-2$ and proving $\sqrt x<x$.

Well, if $x>1$, which is true since $a>3$, then $x^2>x>0$ so $x>\sqrt x$.

That means $a-2>\sqrt{a-2}$ or $b=\sqrt{a-2}+2<a.\quad$ QED

Answer 2

Let $b=f(a)=2+\sqrt{a-2}$, where $a\in (3,5)$. You have proved that $3<f(a)=b$. Now consider $$g(a)=a-f(a)=a-2-\sqrt{a-2}.$$ Then $g(a)$ is a increasing function since $g'(a)=1-\frac{1}{2\sqrt{a-2}}>1-\frac{1}{2\sqrt{3-2}}=\frac{1}{2}>0$. So $$a-b=a-f(a)=g(a)> g(3)= 3-2-\sqrt{3-2}=0.$$

More precisely, we have $0 < a-b < g(5)=5-2-\sqrt{5-2}=3-\sqrt{3}.$