$c\leq xy$ is not a convex condition.


However we know $c\leq xy$ is convex in domain $x,y>0$ in $\mathbb R^2$ for any fixed $c\in\mathbb R$.



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> Is $c\leq x\_1y\_1+\dots+x\_ny\_n$ with $0\leq x\_1,\dots,x\_n\leq 1$and $0\leq y\_1,\dots,y\_n\leq n-1$ a convex condition in $\mathbb R^{2n}$ for any fixed $c\in\mathbb R$?
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Essentialy I have one constraint of form 
$$c\leq x\_1y\_1+\dots+x\_ny\_n$$
$$0\leq x\_1,\dots,x\_n\leq1$$
$$0\leq y\_1,\dots,y\_n\leq n-1$$
and remaining constraints of form $$AX\leq b$$ where $A\in\mathbb R^{m\times 2n}$, $X=\begin{bmatrix}x\_1,\dots,x\_n,y\_1,\dots,y\_n\end{bmatrix}'$ and $b\in\mathbb R^m$ where $m=n^\alpha$ at some fixed $\alpha>0$.



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> Is this quadratic program with single quadratic constraint and polynomially many linear constraints solvable in polynomial time with some known algorithm?
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Linear program with one quadratic condition convex in domain of interest polynomial time solvable?
Even for $n=2$ the constraint is not convex. Note that at $(x\_1,x\_2,y\_1,y\_2) = (0,1,0,1)$ and $(x\_1,x\_2,y\_1,y\_2) = (1,0,1,0)$ we have $x\_1 y\_1 + x\_2 y\_2 =1$,
but at their midpoint $(1/2,1/2,1/2,1/2)$ we have only $x\_1 y\_1 + x\_2 y\_2 = 1/2$. 


