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

Question

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!

Answer 1

If $a,b\ne 0$, then $(a+b)\left(\frac1{a}+\frac{4}{b}\right)=9\iff(a+b)(b+4a)=9ab\iff(2a-b)^2=0$.