proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$ 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
 A: 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.
A: 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.
A: 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 $0 < t < 1$.  
