Minimum value of $\frac{a^2 + b^2 +1 }{a(b+1)}$

Question

Find the minimum value of $\dfrac{a^2 + b^2 +1 }{a(b+1)}$ for $a$, $b$ positive reals.

I know that the solution is $\sqrt{2}$ using perfect squares, derivatives, etc. However, I want to find the minimum using inequalities.

Answer 1

Observe, $$b^2+1\ge \frac{(b+1)^2}{2}$$ Using the A.M-G.M inequality, $$a^2+\frac{(b+1)^2}{2}\ge \sqrt{2}\cdot a(b+1)$$ Equality happens at $b=1, a=\sqrt{2}.$

Answer 2

First, substituting $b=1$ gives an upper bound on the value. $$\inf \left\{ \frac{a^2+b^2+1}{a(b+1)} : a>0,b>0\right\} \leq \inf \left\{ \frac{a^2+2}{2a} : a>0\right\} $$ Second, using the AM-GM inequality

$$\frac{a^2+2}{2a}=\frac{a+2/a}{2} \geq \sqrt{a \cdot 2/a}=\sqrt{2}. $$ Combined this gives an upper bound on the minimum of $\sqrt{2}$.