Name for this Normed Space Inequality: $||u + v||^2 \le 2||u||^2 + 2 ||v||^2$?

Question

The title is the question. Some other common inequalities are named - Schwartz, Bessel, Triangle - is this one ?

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Background.

I saw this stated while reading about functional calculus in Hilbert space: Jirí Blank, Pavel Exner, Miloslav Havlícek - Hilbert space operators in quantum physics p.160

My first thought since it is happening in Hilbert space was that it follows immediately as a consequence of the parallelogram law: $||u + v||^2 + ||u - v||^2 = 2||u||^2 + 2 ||v||^2$.
On further reflection it would seem to be true in any normed space
By triangle inequality $||u + v|| \le ||u|| + ||v||$, so,
$||u + v||^2 \le (||u|| + ||v||)^2 = ||u||^2 + ||v||^2 + 2||u||.||v||$ ....(A)
And for any real a, b (like $||u||, ||v||$ ) $0 \le (a - b)^2 = a^2 + b^2 - 2a.b \implies 2a.b \le a^2 + b^2$ ...(B)

So. from (A) and (B),
$||u + v||^2 \le ||u||^2 + ||v||^2 + 2||u||.||v|| \le 2||u||^2 + 2 ||v||^2$

Answer 1

This one has no name, since it follows quite easily from a triangle inequality.

Answer 2

Maybe the root-mean square vs. AM inequality since it's $\displaystyle\,\sqrt{\frac{a^2+b^2}{2}} \ge \frac{a+b}{2}\,$ for $\,a,b \ge 0$.