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

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.

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.