 # How can I prove that: 0<r^2<(r^2+(wl)^2)((1-w^2lc)^2+(wrc)^2) AA c>0 and all the other variables are bigger than zero using the scalar product of the vectors A=(r,wl) and B=(1-w^2 lc,wrc)? Jackson Garner 2022-09-16 Answered
How can I prove that:
$\begin{array}{}\text{(1)}& 0<{r}^{2}<\left({r}^{2}+\left(wl{\right)}^{2}\right)\left(\left(1-{w}^{2}lc{\right)}^{2}+\left(wrc{\right)}^{2}\right)\end{array}$
$\mathrm{\forall }c>0$ and all the other variables are bigger than zero using the scalar product of the vectors $A=\left(r,wl\right)$ and $B=\left(1-{w}^{2}lc,wrc\right)$?
I do not know how to get started on this problem and I do not see how I can use the scalar product of vectors to prove this inequality.
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For $r\ne 0$ , ${r}^{2}>0$ is trivial. The other inequality is $\left(A\cdot B{\right)}^{2}<\left(A\cdot A\right)\left(B\cdot B\right)$, which is Cauchy–Schwarz for non-parallel vectors (otherwise we could only claim $\le$). In particular, check that $r\left(1-{w}^{2}lc\right)+wl\left(wrc\right)=r$. But A could be parallel to B, so < is in general incorrect.

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