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Let $H$ be a Hilbert space. Is it true that, if $\|x\|$ is less than or equal to $r$ and $\|y\|$ is strictly greater than $r$, then $\left\| x-\frac{ry}{\|y\|} \right\|$ is less than or equal to $\|x-y\|$??

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Hint: Consider linear span of $x$ and $y$, then you reduce your problem to the usual $\mathbb{R}^2$ case.

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For convenience let us write $ry / \|y\| =:\hat{y}$. Note that $y - \hat{y} = \left(1 - r / \|y\|\right) y =: \alpha y$. Note that by assumption $\alpha > 0$.

We have $$ \|x - \hat{y}\|^2 = \|x\|^2 + \|\hat{y}\|^2 - 2 \langle x,\hat{y}\rangle $$

We have $$ \begin{align} \|x - y\|^2 &= \|x\|^2 + \|\hat{y} + \alpha y\|^2 - 2\langle x, \hat{y} + \alpha y\rangle \\ & = \|x\|^2 + \|\hat{y}\|^2 + \|\alpha y\|^2 + 2 \langle \hat{y},\alpha y\rangle - 2 \langle x,\hat{y}\rangle - 2 \langle x,\alpha y\rangle\end{align}$$

Claim $\langle \hat{y},y\rangle - \langle x,y\rangle \geq 0$.

Proof: $\langle \hat{y},y\rangle = r \|y\|$ by definition. $\langle x,y\rangle \leq \|x\| \|y\|$ by Cauchy-Schwarz. By assumption $r \geq \|x\|$. q.e.d.

Hence $$ \| x - y\|^2 - \|x - \hat{y}\|^2 \geq \|\alpha y\|^2 > 0 $$ so the answer is "yes".

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