Why does this inequality hold true for any norm $\|a\| \|b\| \lesssim \delta \|a \|^2 + \delta^{-1} \|b\|^2 $ for $ \forall \delta >0$.

Question

$\|a\| \|b\| \lesssim \delta \|a \|^2 + \delta^{-1} \|b\|^2 $ for any $\delta >0$.

Is this true for any kind of norm? Can you show me why?

Answer 1

\begin{align*} \|a\|\|b\|&=(\delta^{1/2})\|a\|(\delta^{-1/2})\|b\|\\ &=\|\delta^{1/2}a\|\|\delta^{-1/2}b\|\\ &\leq2^{-1}(\|\delta^{1/2}a\|^{2}+\|\delta^{-1/2}b\|^{2})\\ &=2^{-1}(\delta\|a\|+\delta^{-1}\|b\|^{2}). \end{align*}

Answer 2

Sure, it's the AM-GM inequality. It has actually nothing to do with the underlying norm, or any norm for that matter.

Consider $\alpha\stackrel{\rm def}{=} \sqrt{\delta}\lVert a\rVert$ and $\beta\stackrel{\rm def}{=} \sqrt{\delta^{-1}}\lVert b\rVert$. Then, the AM-GM inequality yields $$ \lVert a\rVert\lVert b\rVert = \alpha\beta \leq \frac{\alpha^2+\beta^2}{2} = \frac{1}{2}\left(\delta\lVert a\rVert^2 + \delta^{-1}\lVert b\rVert^2\right) $$

Answer 3

Hint: $\;\delta^2 \|a\|^2 + \|b\|^2 \color{red}{- 2\delta\|a\|\|b\| + 2\delta\|a\|\|b\|} = \big(\delta\|a\|-\|b\|\big)^2 + 2\delta\|a\|\|b\| \ge 2\delta\|a\|\|b\|\,$.