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

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 $0 < t < 1$.  
