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The real triple $(a,b,c)$ satisfy $a+b+c\leq 0$ and $a^2+b^2+c^2\leq 6$. Evaluate the minimum of $$a^2+2b^2+c^2-6a-8b+6c$$

This is a $10$th grade practice problem. I tried using the completing the square method but it does not work.

$$a^2+2b^2+c^2-6a-8b+6c=(a-3)^2-9+2(b-2)^2-8+(c+3)^2-9=(a-3)^2+2(b-2)^2+(c+3)^2\geq -26$$

But Wolfram Alpha gives the minimum as $-19>-26$. So the value of $-26$ that I got could just be an infimum .

nonuser
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    (sorry, had to edit this but deleted by accident) You've completed the square correctly; but what that tells you is that $a^2+2b^2+c^2-6a-8b+6c$ obtains a minimum value of $-26$ when $a=3$, $b=2$ and $c=-3$, which does not satisfy either $a+b+c\le 0$ or $a^2+b^2+c^2\le 6$. What methods do you know (or can you think of) to incorporate these conditions? – Chris Lewis Jan 16 '25 at 11:48
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    @ChrisLewis Maybe geometric presentations could help or Cauchy Schwarz like inequalities ? – nonuser Jan 16 '25 at 11:50
  • Hmmm, $a=1, b=1, c=-2$ should satisfy the equation as -19 (As we clearly should assign $a+b+c=0, a^2+b^2+c^2=6, a>0, b>0, c<0$). Wonder how to prove it though. – Leo Ji Jan 16 '25 at 12:07
  • @LeoJi Tl This is a competition problem ( for the grade 10th) . There is no hint at the end of the book about this particular problem. – nonuser Jan 16 '25 at 12:37

1 Answers1

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Remark. Perhaps this solution is not for 10th grade competition. The approach is similar to my answer to this question. We use completing the square and AM-GM to motivate this solution.

We have \begin{align*} &a^2 + 2b^2 + c^2 - 6a - 8b + 6c - (-19)\\ &\qquad + 2(a + b + c) + (a^2 + b^2 + c^2 - 6)\\ ={}& 2(a - 1)^2 + 3(b-1)^2 + 2(c+2)^2\\ \ge{}& 0. \end{align*} Thus, $a^2 + 2b^2 + c^2 - 6a - 8b + 6c \ge -19$ if $a + b + c \le 0$ and $a^2 + b^2 + c^2 \le 6$.

Also, when $a = 1, b = 1, c = -2$, we have $a^2 + 2b^2 + c^2 - 6a - 8b + 6c = -19$ and $a + b + c = 0$ and $a^2 + b^2 + c^2 = 6$.

Thus, the minimum is $- 19$.

River Li
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