$\sup_{(a,b)\in B_2(0,1)}\left\{\max(|a+2b|,|b+2a|)\right\}=2\sqrt{2}$?

Question

Let $$B_2(0,1):=\{(a,b)\in \mathbb{C}^2;\;|a|^2+|b|^2<1\}.$$

I want to calculate $$N:=\sup_{(a,b)\in B_2(0,1)}\left\{\max(|a+2b|,|b+2a|)\right\}.$$

My attempt: Since for all $(a,b)\in B_2(0,1)$ we have $$\max\{|a+2b|,|b+2a|\}\leq 2(|a|+|b|)\leq 2\sqrt{2}\sqrt{|a|^2+|b|^2}<2\sqrt{2},$$ it follows that $N\leq 2\sqrt{2}$.

It is possible to show that $N=2\sqrt{2}$?

Answer 1

Hint. Note that by Cauchy-Schwarz inequality: $$|a+2b|\leq \sqrt{|a|^2+|b|^2} \cdot \sqrt{1^2+2^2}.$$ Moreover consider for $n\geq 1$, the sequence $(a_n,b_n)=\left(\frac{1}{\sqrt{5+\frac{1}{n}}},\frac{2}{\sqrt{5+\frac{1}{n}}}\right)\in B_2(0,1).$