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Consider the problem \begin{equation} \mbox{min}~~~ x_1^2+2x_2^2+x_1 \\ s.t.~~~ x_1+x_2\le a, \end{equation} where $a\in \mathbb{R}$ is a parameter. I tried to prove it by contradiction by assuming it has two optimal solutions, say, $(x_1,x_2)$ and $(y_1,y_2)$. Then I equated $x_1^2+2x_2^2+x_1=y_1^2+2y_2^2+y_1$ with $x_1+x_2\le a $ and $y_1+y_2\le a$. But from here I am unable to show that $(x_1,x_2)=(y_1,y_2)$. Can anyone please give me a hint or point me in a right direction?

Phoenix8128
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    The function $f(x_1,x_2)=x_1^2+2x_2^2+x_1$ is strictly convex since its Hessian is positive definite. This function is optimised over the convex set $x_1+x_2\leq a$. Strictly convex functions on convex domains have unique minimisers, see https://math.stackexchange.com/questions/337090/if-f-is-strictly-convex-in-a-convex-set-show-it-has-no-more-than-1-minimum . – Levent Feb 21 '23 at 13:34
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    If you want to apply @Levent 's comment to you case, you can show that for any $\lambda\in]0,1[$, $[z_1,z_2]^T=\lambda [x_1,x_2]^T+(1-\lambda) [y_1,y_2]^T$ satisfies the constraint of the problem and has a strictly better score than both points, thus reaching a contradiction. – P. Quinton Feb 21 '23 at 13:44
  • @Levent. Thanks, got it. – Phoenix8128 Feb 21 '23 at 15:44
  • @P. Quinton Yes, I was indeed able to prove with your given $(z_1,z_2)$, the value of objective function is lower. – Phoenix8128 Feb 21 '23 at 15:48

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