Is this inequality true? $u^2+v^2+s^2+t^2\geq (u+v)(s+t)$

Question

Is this inequality true in $\mathbb{R}$? $$u^2+v^2+s^2+t^2\geq (u+v)(s+t)$$

I don't know if this is a well-known result. If you have a counterexample or a relevant reference I would appreciate it.

Answer 1

\begin{eqnarray*} (u-s)^2+ (u-t)^2+(v-s)^2+(v-t)^2\geq 0. \end{eqnarray*} Now divide by $2$ and rearrange.

Answer 2

It is $$(u-v)^2+(v-s)^2+(u-t)^2+(v-t)^2\geq 0$$

Answer 3

Yes, it's true.

By C-S and AM-GM we obtain: $$u^2+v^2+s^2+t^2=\frac{1}{2}\left((1^2+1^2)(u^2+v^2)+(1^2+1^2)(s^2+t^2)\right)\geq$$ $$\geq\frac{1}{2}\left((u+v)^2+(s+t)^2\right)\geq\sqrt{(u+v)^2(s+t)^2}=|(u+v)(s+t)|\geq(u+v)(s+t).$$