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proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$

for $t\leq 1,$ left side expression is $>0$

for $t\geq 1,$ left side expression $t^5(t-1)+t^3(t-1)+t(t-1)+0.4$ is $>0$

i wan,t be able to prove for $0<t<1,$ could some help me with this

DXT
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    This can be written as $\frac{t^7+1}{t+1}-0.6$, if that helps. – Thomas Andrews Dec 29 '16 at 04:45
  • @ThomasAndrews - Nice hint. But how do we know that $(t^7+1)/(t+1)>0.6$ by inspection? – Hypergeometricx Dec 29 '16 at 17:12
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    I don't know - I said "If that helps" because I didn't know if it helps. @hypergeometric – Thomas Andrews Dec 29 '16 at 17:43
  • @ThomasAndrews The minimum of the function is approximately $0.351$. – egreg Dec 30 '16 at 00:55
  • i also tried using arithmetic geometric inequality, $\displaystyle \frac{t^6}{2}+\frac{t^4}{2}\geq t^5;,\frac{t^4}{2}+\frac{t^2}{2}\geq t^3;,\frac{t^2}{2}+\frac{2}{5}\geq \frac{|t|}{5}\geq \frac{t}{5}$ and $\displaystyle \frac{t^6}{2}\geq 0$ but not suceed – DXT Dec 30 '16 at 04:04

3 Answers3

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By breaking the polynomial into groups of three terms and completing the square, we get: \begin{align} & \hspace{0.36 in} t^6-t^5+t^4-t^3+t^2-t+\dfrac{2}{5} \\ &= \left(t^6-t^5+\dfrac{1}{4}t^4\right)+\dfrac{3}{4}t^4-t^3+t^2-t+\dfrac{2}{5} \\ &= \left(t^6-t^5+\dfrac{1}{4}t^4\right)+\left(\dfrac{3}{4}t^4-t^3+\dfrac{1}{3}t^2\right)+\dfrac{2}{3}t^2-t+\dfrac{2}{5} \\ &= \left(t^6-t^5+\dfrac{1}{4}t^4\right)+\left(\dfrac{3}{4}t^4-t^3+\dfrac{1}{3}t^2\right)+\left(\dfrac{2}{3}t^2-t+\dfrac{3}{8}\right)+\dfrac{1}{40} \\ &= t^4\left(t^2-t+\dfrac{1}{4}\right)+\dfrac{3}{4}t^2\left(t^2-\dfrac{4}{3}t+\dfrac{4}{9}\right)+\dfrac{2}{3}\left(t^2-\dfrac{3}{2}t+\dfrac{9}{16}\right)+\dfrac{1}{40} \\ &= t^4\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}t^2\left(t-\dfrac{2}{3}\right)^2+\dfrac{2}{3}\left(t-\dfrac{3}{4}\right)^2+\dfrac{1}{40} \\ &> 0. \end{align}

EDIT: It can be shown via induction that for any $N \in \mathbb{N}$, the following holds: $$\dfrac{N}{2N+2}+\displaystyle\sum_{k = 1}^{2N}(-1)^kt^k = \displaystyle\sum_{n = 1}^{N}\dfrac{n+1}{2n}t^{2(N-n)}\left(t-\dfrac{n}{n+1}\right)^2 \ge 0.$$ Thanks to hypergeometric for suggesting to look into this.

JimmyK4542
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  • Interesting to note that the "completed squares" form a nice pattern: $(t-1/2)^2, ; (t-2/3)^2; (t-3/4)^2$. Is it possible to exploit this a posteriori to derive the last line in one step? – Hypergeometricx Dec 29 '16 at 17:08
  • Nice identity. Wonder what the best possible constant is, since different values of $t$ zero different squares. – marty cohen Dec 30 '16 at 01:08
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Let $p(t) = t^6 - t^5 + t^4 - t^3 + t^2 - t +2/5$. Observe that $$ p(t) = \begin{bmatrix} 1\\t\\t^2\\t^3\end{bmatrix}^\intercal \begin{bmatrix}2/5&-1/2&0&0\\-1/2&1&-1/2&0\\0&-1/2&1&-1/2\\0&0&-1/2&1\end{bmatrix}\begin{bmatrix} 1\\t\\t^2\\t^3\end{bmatrix} $$ The matrix in the middle is positive definite, from which it follows immediately that $p(t) > 0$ for all $t$.

Edit: Positive definiteness can be determined by mechanically calculating the matrix's minors, which is easy.

Henricus V.
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If $0< t < 1$

$\frac {t^7 + 1}{t+1} \ge .6 \iff t^7 + 1 \ge .6t + .6 \iff t^7 - .6t \ge -.4$

$\frac {d(t^7 - .6t)}{dt} = 7t^6 - .6 = 0$ if $t = \sqrt[6] \frac 6{70}$

$\frac{d^2(t^7 - .6t)}{d^2t} = 42t^5 > 0 $ if $t > 0$ so $t = \sqrt[6] \frac 6{70}$ is a minimum value of $t^7 - .6t$. And so $t^7 - .6t \ge \sqrt[6] \frac 6{70}^7 -.6*\sqrt[6] \frac 6{70} = \sqrt[6] \frac 6{70}(\frac 6{70} - .6)\approx -.341 > -.4$

so $t^7 - .6t \ge -.4$ for $0 < t < 1$.

So $\frac {t^7 + 1}{t+1}= t^6 - t^5 + t^4 - t^3 +t^2 -t + 1 \ge .6$

and $t^6 - t^5 + t^4 - t^3 +t^2 -t + .4 > 0$ for $0 < t < 1$.

fleablood
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