Let $x \in \mathbb{R}^3$ and $$p(x)=x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_1x_3. $$ For any $a \in \mathbb{R}^3$, show that $$H_a={x\in \mathbb{R}^3: p(x) +a \cdot x+1 <0}$$ convex.
Please help me to solve the above problem. Thank you.
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What have you tried? Where are you stuck? – Igor Rivin Mar 28 '21 at 14:21
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The set $H_a$ is convex if for any $x, y \in \mathbb{R} ^3$ and $t\in [0,1]$, $$p(tx+(1-t)y)+ a\cdot (tx+(1-t)y)+1<0$$. How to show this inequality holds? – flourence Mar 28 '21 at 14:23
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Hint
Show that $p(x)+a\cdot x+1$ is convex by finding its Hessian matrix and showing it to be positive semi-definite.
Mostafa Ayaz
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