Ratio of polynomials, how to prove $f(t) \ge 0$ for $t > 0$? I am looking at the following function $$
f(t) = \frac{t+1}{t^2} - \frac{16}{(t+1)^3}
$$
and am struggling to prove that $f(t) \ge 0$ whenever $t>0$. The statement appears true from plotting and inspecting the graph, but this is far from a proof.
 A: Combine the fractions and keep factoring
$$\frac{t+1}{t^2}-\frac{16}{(t+1)^3} = \frac{(t+1)^4-16t^2}{t^2(t+1)^3} = \frac{[(t+1)^2-4t][(t+1)^2+4t]}{t^2(t+1)^3}$$
$$ = \frac{(t-1)^2[(t+1)^2+4t]}{t^2(t+1)^3}$$
which is always nonnegative for $t>0$ (and $0$ when $t=1$)
A: What the OP was trying to prove (prior to an edit) was not exactly true.
For $t\gt0$, we have
$$\begin{align}
{t+1\over t^2}\gt{16\over(t+1)^3}
&\iff(t+1)^4\gt16t^2&\text{(clearing denominators)}\\
&\iff(t+1)^2\gt4t&\text{(taking square roots)}\\
&\iff t+1\gt2\sqrt t&\text{(taking sqare roots again)}\\
&\iff t-2\sqrt t+1\gt0&\text{(simple algebra)}\\
&\iff(\sqrt t-1)^2\gt0&\text{(more simple algebra)}
\end{align}$$
The final inequality is true if $t\not=1$ but false if $t=1$. However, if (as the OP did) we replace the strict inequality sign $\gt$ with non-strict inequality sign $\ge$ in the display, then things are true for all $t\gt0$.
A: We have that
$$f(t) = \frac{t+1}{t^2} - \frac{16}{(t+1)^3} =\frac{t^4+4t^3-10t^2+4t+1}{t^2(t+1)^3}$$
the denominator is positive the we can consider the sign of $p(t)=t^4+4t^3-10t^2+4t+1$ and we have
$$p'(t)=4t^3+12t^2-20t+4 =0 \implies p(1)=0$$
then we find
$$p'(t)=(t-1)(t^2+4t-1)$$ and
$$t^2+4t-1=0 \implies t=\frac{-4\pm\sqrt{16+4}}{2}=-2\pm \sqrt 5$$
an inspection to $p''(t)$ reveals that $t=1$ is a local minimum and $t=-2+ \sqrt 5$ is a local maximum and we have
$$p(1) =0$$
therefore for $t>0$ we have $f(t)\ge 0$ with $f(t)=0 \iff t=1$.
