Find the minimum value $2a^8+2b^6+a^4-b^3-2a^2-2$, where $a$ and $b$ are real numbers. I was told to use Lagrange multipliers but I found out this belongs to Calculus department. I tried factoring the expression but it's irreducible. Is there really a way to solve by algebra?
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We have $$2a^8+2b^6+a^4-b^3-2a^2-2$$$$=(2a^8+a^4-2a^2)+(2b^6-b^3)-2$$$$=\left(2\left(a^4-\frac 14\right)^2+2\left(a^2-\frac 12\right)^2-\frac 58\right)+\left(2\left(b^3-\frac 14\right)^2-\frac 18\right)-2$$ $$=2\left(a^4-\frac 14\right)^2+2\left(a^2-\frac 12\right)^2+2\left(b^3-\frac 14\right)^2-\frac{11}{4}.$$ Hence, the minimum value is $-\frac{11}{4}$ when $(a,b)=\left(\pm\frac{1}{\sqrt 2},\sqrt[3]{\frac{1}{4}}\right).$
mathlove
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wow! ur so amazing! thanks a lot. – keith_here Sep 23 '14 at 13:14
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@keith_here: You are welcome. – mathlove Sep 23 '14 at 14:27