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Given $A$ symmetric and semidefinite positive, for each $x$ $$ x'Ax \geq \frac{1}{\Vert A\Vert} \Vert Ax \Vert^2 $$

This inequality appears at page 24 of "Introduction to Optimization" from Boris T. Polyak. I haven't been able to prove it. Any idea?

Thanks, Giovanni

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1 Answer 1

up vote 3 down vote accepted

The claim is that $|A\| x' A x \ge \|A x\|^2 = x' A^2 x$, i.e. that $\|A\| A \ge A^2$. This comes from $\|A\| I \ge A$ by multiplying on the left and right by the positive semidefinite square root of $A$.

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