# $\frac{\alpha}{\alpha + \beta}\|u\|^2 + \frac{\beta}{\alpha + \beta}\|v\|^2 > \langle u, v\rangle$ is true?

I have the following question:

It's true that $$\frac{\alpha}{\alpha + \beta}\|u\|^2 + \frac{\beta}{\alpha + \beta} \|v\|^2 > \langle u, v\rangle$$

for all $u,v \in \mathbb R^N$ with $\{u, v\}$ linearly independent and all $\alpha, \beta \in (0, \infty)$?

where $\langle \cdot, \cdot \rangle$ is the usual inner product in $\mathbb R^N$ and $\|\cdot\|$ is the euclidean norm.

In the particular case that $\alpha = \beta = \frac{1}{2}$ is easy. If $\{u, v\}$ linearly independent then by the Hölder inequality $\frac{1}{2}\|u\|^2 + \frac{1}{2}\|v\|^2 > \|u\|\cdot\|v\| \geq |\langle u, v\rangle| \geq \langle u, v\rangle$. but in the general case I do not know how to do.

• If $\alpha=0$ and $\beta>0$, the inequality is $||v||^2>\langle u, v\rangle$ which is not true for all $u,v$. – yurnero May 14 '18 at 2:03
• also when $u=v$ the strict inequality cannot hold – zwim May 14 '18 at 2:07
• Thank you so much. I will put the condition $\alpha$ and $\beta$ strictly positive. And $u \neq v$. – Luiz Collovini May 14 '18 at 2:10
• @LuizCollovini That's doesn't fix it. As you can fix $\beta>0$ say ($\beta=1$) and let $\alpha\downarrow 0$, which implies $||v||^2\geq \langle u, v\rangle$ which still isn't true for all $u\neq v$. – yurnero May 14 '18 at 2:18
• @LuizCollovini If it is not a typing error, note that your argument works for any $\alpha = \beta$ (it is not necessary to be restricted to $\alpha = \beta = \frac{1}{2}$). – Pedro May 14 '18 at 2:46

Unfortunately that is not true. What are you asking is equivalent to \begin{align} \frac{\alpha}{\alpha + \beta}\|u\|^2 + \frac{\beta}{\alpha + \beta} \|v\|^2 &> \langle u, v\rangle \\ \alpha\|u\|^2 + \beta \|v\|^2 &> (\alpha + \beta)\langle u, v\rangle \\ \alpha\langle u, u\rangle - (\alpha + \beta)\langle u, v\rangle + \beta \langle v, v\rangle &> 0 \\ \langle \alpha u - \beta v, u - v\rangle &> 0 \end{align}
Now since $u$ and $v$ are arbitrary we can drop the minus sign, thus obtaining $$\langle \alpha u + \beta v, u + v\rangle > 0$$
To see that the above inequality does not hold, in $\mathbb{R}^2$, take $u = (2, -\frac{1}{2})$, $v = (-1, \frac{1}{2})$, $\alpha = 1$ and $\beta = 3$. Hence $\alpha u + \beta v = (-1, 1)$ and $u + v = (1,0)$, which clearly does not satisfy the inequality.