If $H$ is positive definite and $s^Ty>0$, then $s^THs-\frac{s^Tyy^Ts}{s^Ty+y^TH^{-1}y}\ne -1$

Question

Let

• $H\in\mathbb{R}^{n\times n}$ be symmetric and positive definite
• $s,y\in\mathbb{R}^n$ with $s^Ty>0$

How can we show, that $$s^THs-\frac{s^Tyy^Ts}{s^Ty+y^TH^{-1}y}\ne -1\;?\tag{1}$$

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Clearly, $H^{-1}$ is symmetric and positive definite, too, and since $s^Ty>0$, we must have $s,y\ne 0$. So $$s^THs>0\;,\;\;\;s^Tyy^Ts>0\;,\;\;\;y^TH^{-1}y>0\;.$$ However, I still don't see how we can conclude, that $(1)$ must be true.

Answer 1

$||.||$ denotes the Euclidean norm and $s^Ty=(s,y)$ denotes the standard scalar product. Note that $signum(s^THs-\dfrac{(s,y)^2}{(s,y)+y^TH^{-1}y})=signum((s^THs)(s,y)+(s^THs)(y^TH^{-1}y)-(s,y)^2)$. It suffices to show that $f=(s^THs)(y^TH^{-1}y)-(s,y)^2\geq 0$. Now $f=||H^{1/2}s||^2||H^{-1/2}y||^2-(H^{1/2}s,H^{-1/2}y)^2$ and we conclude using Cauchy-Schwarz.