<s>[INST] 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!
 [/INST] 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: $\dfrac{1}{a}+\dfrac{4}{b} = \dfrac{1^2}{a}+\dfrac{2^2}{b} \ge \dfrac{(1+2)^2}{a+b} = \dfrac{9}{a+b}$.
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