Prove $2a^Tb \leq \|a \|^2 + \|b\|_*^2$ with dual norm

Question

We know we can easily (**) prove $$ 2a^Tb \leq \|a \|_2^2 + \|b\|_2^2, \forall a,b $$ Is there a way to prove the following: $$ 2a^Tb \leq \|a \|^2 + \|b\|_*^2, \forall a,b $$

where $\|\cdot\|_*$ is the dual norm of $\|\cdot\|$.

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(**) Proof:
$$ \| a - b \|_2^2 =(a-b)^T (a-b) = \|a \|_2^2 + \|b\|_2^2 - 2a^Tb \geq 0 $$

Answer 1

By definition, of the dual norm, $$ a^Tb\leq \|a\|\|b\|_*. $$ Then, using the arithmetic-geometric inequality (or $2xy\leq x^2+y^2$ if you want), $$ 2a^Tb\leq 2\|a\|\|b\|_*\leq \|a\|^2+\|b\|_*^2. $$