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Let $a,b,c$ be reals such that $a-b+c=3, a^2+b^2+c^2=4$. Find the maximum and minimum of $\sqrt{2}a + \sqrt{2}b + 3c$.

Don't use coordinates or Lagrange multipliers in order to solve this. Are there any algebraic ways to solve this that don't involve Calculus?

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I saw this problem in AoPS but didn't solve it because it was bashy. Anyways..

If Lagrange Multipliers is banned, what we can do is set $\sqrt{2}a+\sqrt{2}b+3c=k$, and with $a-b+c=3$, we can write $$a=\frac{k+3\sqrt{2}-(3+\sqrt{2})c}{2\sqrt{2}}, b=-3+\frac{k+3\sqrt{2}-(3+\sqrt{2})c}{2\sqrt{2}}+c$$

Plug these values into $a^2+b^2+c^2=4$. You will have a quadratic (very dirty) in $c,k$.

Take the determinant to find the range of $k$.

Solving gives the minimum as $-0.1622$ and maximum as $6.1622$.

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