Prove that $\max_x \frac{(x^Tv)^2}{x^TBx}=v^TB^{-1}v$

Question

Let $B>0$ be positive definite $d \times d$ matrix, and $v \in \mathbb R^d$. Prove that $$\max_x \frac{(x^Tv)^2}{x^TBx}=v^TB^{-1}v$$ Hint: Change of variable $z=B^{1/2}x$

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$$\begin{split}\max_z\frac{(v^TB^{-1/2}z)^2}{z^Tz}\end{split}$$

Take the derivative

$$\begin{split}\frac{(z^Tz)2(v^TB^{-\frac 1 2}z)B^{-\frac 1 2}v-(v^TB^{-\frac 1 2}z)^2(2z)}{(z^Tz)^2} &=\frac{2(z^Tz)(v^TB^{-\frac 1 2}z)B^{-\frac 1 2}v-2(v^TB^{-\frac 1 2}z)(v^TB^{-\frac 1 2}z)z}{z^Tzz^Tz}\\ &=\frac{2(z^T\not{z)(z^T}B^{-\frac 1 2}v)B^{-\frac 1 2}v-2(v^TB^{-\frac 1 2}\not{z)(z^T}B^{-\frac 1 2}v)z}{z^T\not{zz^T}z}\\ &=\frac{2(z^TB^{-\frac 1 2}v)B^{-\frac 1 2}v-2(v^TB^{-1}v)z}{z^Tz}\end{split}$$

Probably the last cancellation is invalid, wasn't sure what to do, or how to proceed.

Answer 1

Following Semiclassical's comment, it suffices to show $\max_z \frac{(u^\top z)^2}{z^\top z} = u^\top u$. The Cauchy-Schwarz inequality implies $\max_z \frac{(u^\top z)^2}{z^\top z} \le u^\top u$. To show equality is attained, take $z=u$ (or $z=cu$ for any scalar $c$).