$\ \|u\|_{H^{s}} \leq\|u\|_{s_{0}}^{\alpha}\|u\|_{s_{1}}^{1-\alpha}, \quad \forall u \in H^{s_{1}}\left(\mathbb{R}^{n}\right) $

Question

I'm completely in the dark about this problem let $s_{0}<s_{1}$ and $s=\alpha s_{0}+(1-\alpha) s_{1},$ where $\alpha \in [0,1]$

The question is to proof that $$\ \|u\|_{H^{s}} \leq\|u\|_{s_{0}}^{\alpha}\|u\|_{s_{1}}^{1-\alpha}, \quad \forall u \in H^{s_{1}}\left(\mathbb{R}^{n}\right) $$

where: $H^{s}\left(\mathbf{R}^{N}\right)=\left\{u \in \mathcal{S}^{\prime}\left(\mathbf{R}^{N}\right)\left|\left(1+|\xi|^{2}\right)^{s / 2} \mathcal{F} u \in L^{2}\left(\mathbf{R}^{N}\right)\right\}\right.$

and $\|u\|_{H^{s}}=\left(\int_{\mathbb{R}^{n}}\left(1+\|\xi\|^{2}\right)^{s}|\widehat{u}(\xi)|^{2} d \xi\right)^{\frac{1}{2}}, \quad u \in H^{s}\left(\mathbb{R}^{n}\right)$

I started as folow:

$\left(1+\|\xi\|^{2}\right)^{s}|\widehat{u}(\xi)|^{2}=\left(1+\|\xi\|^{2}\right)^{\alpha s_{0}}|\widehat{u}(\xi)|^{2 \alpha}\left(1+\|\xi\|^{2}\right)^{(1-\alpha) s_{1}}|\widehat{u}(\xi)|^{2(1-\alpha)}$

$=[\left(1+\|\xi\|^{2}\right)^{ s_{0}}|\widehat{u}(\xi)|^{2 }]^{\alpha}[\left(1+\|\xi\|^{2}\right)^{s_{1}}|\widehat{u}(\xi)|^{2}]^{(1-\alpha)}$

for me, it's clear that Holder inequality is the next step, but according to the norm previously defined, I can't see how. [EDIT: I hope that someone has seen this inequality before]