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

For the inequality your method is correct, but when equality occurs is .. Well, one must note that Cauchy-Schwarz is AM-GM in disguise: $$(a^2+b^2)(c^2+d^2) \ge (ac+bd)^2 \iff a^2d^2+b^2c^2 \ge 2abcd$$ And as in AM-GM, equality occurs if $ad=bc$, applying this to your case we get $$\sqrt{\frac{4a}{b}}=\sqrt{\frac{b}{a}} \iff 2a=b$$

Answer 2

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

Answer 3

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: $\dfrac{1}{a}+\dfrac{4}{b} = \dfrac{1^2}{a}+\dfrac{2^2}{b} \ge \dfrac{(1+2)^2}{a+b} = \dfrac{9}{a+b}$.