# 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\$ Answer

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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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