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I was reading a proof that rewrote an inequality in the form:

$b^Tx +x^T A x \le \alpha$

for $b,x \in \mathbb{R}^n$ and $\alpha \in \mathbb{R}$, and with $A$ positive semidefinite. It then concluded that the solution set is convex. Why is this the case? I can see that $b^Tx$ is an affine function in $x$ and the latter is some sort of ellipsoid? But I'm not sure why their sum in the inequality would lead one to conclude so readily that the solution set is convex?

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Note that both $b^Tx$ and $x^TAx$ are convex functions, and that the sum of convex functions is convex, thus $b^Tx+x^TAx$ is convex. It is a fact that the sublevel sets of a convex function $f$, i.e. the sets $\{x:f(x)\le \alpha\}$, are convex.

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    @PalaceChan The Hessian *is* $A$. Note that $x^TAx=\sum\limits_{i,j} a_{ij}x_ix_j$ so the partial derivatives are just $a_{ij}$.2012-09-27