Upper Bound on scalar product

Question

I know that the following identity holds for $a, b \in \mathbb{R}^d$ and $\alpha > 0$ \begin{align} 2 \langle a, b \rangle \leq \alpha \|a\|^2 + \frac{1}{\alpha} \|b\|^2 \end{align} My question is if I have two sequences of vectors $a_k, b_k \in \mathbb{R}^d$, can I say that \begin{align} 2 \langle a_k, b_k \rangle \leq \alpha_k \|a_k\|^2 + \frac{1}{\alpha_k} \|b_k\|^2 \end{align} where $\alpha_k$ is a strictly positive sequence, i.e. will the inequality sill holds if $\alpha$ depends on $k$?

Answer 1

This comes from Cauchy-Schwartz and Young's inequality \begin{align} 2\langle a,b \rangle \leq 2||a||\cdot||b|| =& 2||\sqrt{\alpha}a||\cdot||\frac{1}{\sqrt{\alpha}}b||\\ \leq &2 \left ( \alpha\frac{||a||^2}{2} + \frac{1}{\alpha}\frac{||b||^2}{2}\right). \end{align}

Cauchy-Schwartz holds for any elements in the inner product space, and Young's for any positive numbers, so if you make a new sequence $ a_k' = \sqrt{\alpha_k} a_k, b_k' = \frac{1}{\sqrt{\alpha_k}}b_k $, then \begin{align} \langle a_k,b_k \rangle = \langle a_k',b_k' \rangle \leq ||a_k'||^2 +||b_k'||^2 \end{align} by the same argument.