Prove $(a+b)\left(\frac{1}{a}+\frac{4}{b}\right)\geq9$ with $a>0$ and $b > 0$. When does equality hold? 
Prove that $(a+b)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)\geq9$ with $a>0$ and $b > 0$. When does equality hold?

My attempt:
By Cauchy-Schwarz inequality, we have:
$$\begin{align*}(a+b)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)&\geq{\left(\sqrt{a\cdot\dfrac{1}{a}}+\sqrt{b\cdot\dfrac{4}{b}}\right)}^2\\&={\left(\sqrt{1}+\sqrt{4}\right)}^2\\&={\left(1+2\right)}^2\\&=(3)^2\\&=9\end{align*}$$
Is it correct?
I'm having trouble showing when equality holds. Could someone help me to understand?
Thanks in advance!

They called it the Titu's lemma in the inequality world or the Angel form of the CS inequality. The inequality follows from this lemma: $\dfrac{1}{a}+\dfrac{4}{b} = \dfrac{1^2}{a}+\dfrac{2^2}{b} \ge \dfrac{(1+2)^2}{a+b} = \dfrac{9}{a+b}$.
