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If $x^5 - x^3 + x = a. $ Then we have to find the minimum value of $x^6$ in terms of a. The answer given is $2a - 1$ if that gives any idea.

I have no idea how to approach this problem. A hint would do fine.

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    $\begingroup$ No other conditions? So if $a=0$ we should expect $x^6=-1$?? to be the minimum? I would be quite sure that $0$ would be the minimum then. $\endgroup$ – Macavity Dec 15 '15 at 12:56
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    $\begingroup$ @Macavity I think it's like the lower bound for $x^6$. $\endgroup$ – Sudhanshu Dec 15 '15 at 12:59
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    $\begingroup$ There is considerable difference between calling something a minimum and lower bound. I thought you wanted a minimum, which means there must be some value of $x \in \mathbb R$ which attains the value $2a-1$. $\endgroup$ – Macavity Dec 15 '15 at 13:00
  • $\begingroup$ @Macavity Sorry then I really don't know and there's nothing else given in the question. $\endgroup$ – Sudhanshu Dec 15 '15 at 13:02
  • $\begingroup$ Express $x^6 = x(x^5 - x^3 + x) + x(x^3 - x) = ax + x^4 - x^2$ $\endgroup$ – Shailesh Dec 15 '15 at 13:07
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I reinterprete the problem a bit differently (based upon the proposed answer) to show that $$\tag1x^5-x^3+x=a\implies x^6\ge 2a-1. $$ Note that $(1)$ is trivially true for $a\le \frac12$. Hence we may assume that $a$ is positive. We have $$(x^2+1)a=(x^2+1)(x^5-x^3+x)=x^7+x $$ so that $x$ must be positive. Divide by $x$ and subtract $1$ to arrive at $$x^6=\frac{(x^2+1)a}{x}-1\ge\frac{2xa}{x}-1=2a-1, $$ where we used $x>0$ and $x^2+1\ge 2x$ (from $x^2+1-2x=(x-1)^2\ge 0$).

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Hint:

Minimizing $x^6$ is the same as finding when $|x|$ is a minimum. Hence we are looking for the real root with smallest absolute value, of $x^5-x^3+x = a$.

Now $x^5-x^3+x $ is strictly increasing, so for any value of $a$, there is only one real root (say $\alpha$) for the polynomial, and hence the minimum is simply $\alpha^6$.

Some cases are easy: $\alpha = a $ when $a \in \{0, \pm1\}$ for e.g. Finding $\alpha$ as a simple function of $a$ does not seem feasible in general though.

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