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- Feb 14, 2012

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I have encountered an interesting math problem which I couldn't solve.

Problem:

Find all values of such that $x^6-6x^5+12x^4+ax^3+12x^2-6x+1$ is non-negative for all real .

1st attempt:

I first let $f(x)=x^6-6x^5+12x^4+ax^3+12x^2-6x+1$

Then if it has repeated roots (of multiplicity of 2 or 4), then the function of f will always greater than zero and we are done.

This leads to a few cases to be considered:

1. $f(x)=(x-p)^4(x-q)^2$

2. $f(x)=(x-p)^2(x-q)^2(x-r)^2$

3. $f(x)=(x-p)^2(\text{always positive})$

4. $f(x)=(x-p)^4(\text{always positive})$

But I notice even if I compare the coefficients of the $x^n$ terms, I can get nothing useful from this attempt and hence, the possible values of a remain unknown.

2nd attempt:

I noticed $\dfrac{x^6+1}{2} \ge x^3$, $\dfrac{6x^5+6x}{2} \ge 6x^3$ and hence $-\dfrac{6x^5+6x}{2} \le -6x^3$, $\dfrac{12x^4+12x^2}{2} \ge 12x^3$.

So, by putting them together, $x^6-6x^5+12x^4+ax^3+12x^2-6x+1 \ge2x^3-12x^3+24x^3+ax^3>=26x^3+ax^3$

If we want $f(x)\ge 0$, then no conclusion can be drawn from this silly attempt.

3rd attempt:

I noticed f(x) could be rewritten as $x^6-6x^5+12x^4+ax^3+12x^2-6x+1=x^4(x^2-6x+12)+ax^3+12x^2-6x+1$

That is, $x^2-6x+12$ and $12x^2-6x+1$ are both quadratic functions with the nice changing of the coefficients, but I don't know what that indicates...perhaps I'm not thinking enough but I will and will add to this thread if I think of something useful from this.

Could someone please point me with the correct direction so that I can get this problem solved?