A Finnish mathematics competition asked to prove that for all $x$ we have $x^8-x^7+2x^6-2x^5+3x^4-3x^3+4x^2-4x+\frac{5}{2}\geq 0$ for all real $x$. I heard that it follows from Hilbert's problem that one can prove this by writing the polynomial as sum of squares. How can I find such a representation? I managed to prove the inequality by considering cases $x\leq 0$, $0
Proving that a polynomial is positive
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0One observation: your polynomial is equal to $x(x-1)(x^6+2x^4+3x^2+4)+2.5$. I'm not sure if this is the path to take, though. – 2011-11-19
2 Answers
I'm not sure if this is what you're looking for, but it works and is quite elementary.
We can take the highest three powers and do this: $ x^8-x^7+x^6 = x^6(x^2-x+1) = x^6\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right] = x^6 \left(x-\dfrac{1}{2}\right)^2 + \dfrac{3}{4} x^6, $ which gives us the expression $ x^6 \left(x-\dfrac{1}{2}\right)^2 + \dfrac{3}{4} x^6 + −2x^5+3x^4−3x^3+4x^2−4x + \dfrac{5}{2}. $ Doing this three more times results in $ x^6 \left(x-\dfrac{1}{2}\right)^2 + \dfrac{3}{4}x^4\left(x-\dfrac{4}{3}\right)^2 + \dfrac{5}{3}x^2\left(x-\dfrac{9}{10}\right)^2 + \dfrac{53}{20}\left(x-\dfrac{40}{53}\right)^2 + \dfrac{105}{106} \geq 0. $
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0Yes, you're right. It will only work in certain cases. – 2011-11-19
Dane commented "One observation: your polynomial is equal to $x(x-1)(x^6+2x^4+3x^2+4)+2.5$. I'm not sure if this is the path to take, though."
Yes it is. Notice that when x(x-1)>= 0, the polynomial is positive. If x(x-1) < 0, then 0 -1/4, and 0 < x^6 + 2x^4 + 3x^2 + 4 < 10. The polynomial > -1/4*10 + 2.5=0.
I guess this is the "official" answer.