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pr.chance – If $x ge 0$ and $mathbf{1}^Tx le |x|^2$ then $mathbf{1}^T(I – xx^T / |x|^2) mathbf{1} ge | [mathbf{1} – x]_+ |^2$

**Notation.** Denote $mathbf{1}=(1,1,ldots,1)$ because the vector-of-ones in $mathbb{R}^n$. Write the “positive part” as $[alpha]_+ = max{alpha,0}$ for $alphainmathbb{R}$ and $[(x_1,x_2,ldots,x_n)]_+ = ([x_1]_+,[x_2]_+,ldots,[x_n]_+)$ for $xinmathbb{R}^n$. Example: $[(1,-2,0,3)]_+ = (1,0,0,3)$.

Let $x in mathbb{R}^n$ answer $x ge 0$ and $mathbf{1}^Tx le |x|^2$ the place nonnegativity is interpreted elementwise. Then, I surmise the next lower-bound $$mathbf{1}^T(I – xx^T / |x|^2) mathbf{1} ge | [mathbf{1} – x]_+ |^2$$

Why is that this certain undoubted? (Or is there a counterexample?)

I’ve spent hours verifying the certain over random $x$ and couldn’t discover a counterexample.

Note that the pretense is disloyal if we supplant the RHS by $| mathbf{1} – x |^2$; for instance, let $x=(0,2)$. I believe {that a} proof should in some way labor with the cardinality of $x$, since $mathbf{1}^Tmathbf{1}=n$ and $mathbf{1}^Tx=|x|_1$ and $(|x|_1/|x|_2)^2lemathrm{card}(x)$.

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