Proofs involving strict inequalities 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 $b \gt 2$

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 $0 < a-b < g(5)=5-2-\sqrt{5-2}=3-\sqrt{3}.$
