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

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.
