Two inequalities related to norm We have some difficulties in the following problem:
Let $H$ be a real Hilbert space.


*

*Find $\alpha>0$ such that
$$
\langle\frac{u}{\sqrt{\|u\|}}-\frac{v}{\sqrt{\|v\|}}, u-v\rangle\geq 
\alpha\|u-v\|^2
$$
for all $u, v\in H\setminus\{0\}$ and $\|u\|\leq 1, \|v\|\leq 1$.

*Find $\beta>0$ such that
$$
\langle\frac{u}{\sqrt{\|u\|}}-v, u-v\rangle\geq
\beta\|u-v\|^2
$$
for all $u, v\in H\setminus\{0\}$ and $\|u\|\leq 1, \|v\|\geq 1$.
Thank you for all comments and kind help.
 A: For the first problem, let $u,v = t^2 x, s^2 y$ where $t,s \in [0,1]$ are reals and $x,y$ are unit vectors, and also $\gamma := \left<x,y\right> \in [-1,1 ]$. Then, the expressions on either side become $t^3+s^3 -st(s+t) \gamma$ and $\alpha(t^4 + s^4 - 2 t^2s^2 \gamma)$. What you are looking for is $$\alpha_0 := \min_{t,s,\gamma}{\frac{t^3+s^3 -st(s+t) \gamma}{t^4 + s^4 - 2 t^2s^2 \gamma}}$$ (then, all $\alpha \leq \alpha_0$ will work; note that from $t,s,\gamma$ one can reconstruct some appropriate $u,v$). It is not hard to see that the expression to minimise is monotoneous in $\gamma$, so one can always assume that $\gamma$ takes the extreme values $\gamma \in \{-1,+1\}$. For $\gamma = +1$, the expression becomes just $\frac{1}{t+s}$, so the value is minimal for $t=s =1$, and is $\alpha_+ := \frac{1}{2}$. For $\gamma = -1$, the expression also simplifies, this time to $\frac{t+s}{t^2 +s^2}$, and is minimised when $\{s,t\} = \{0,1\}$ with value $\alpha_{-} := 1$ (for a proof, by scaling you can notice that one of $s,t$ has to be $1$ at the minimum, so you are just minimising $\frac{1+t}{1+t^2}$, and this is just one variable, so can be done with a little work). Finally, $\alpha_0 = \min\{\alpha_+,\alpha_-\} = \frac{1}{2}$, so the bottom line is: any $\alpha \leq \frac{1}{2}$ will do.
For the other problem, I believe you can have a similar reasoning. I have not worked out the details.
