Proof that $ \int_0^\infty \frac{(f'(x))^2}{f(x)^{1980}} dx < \infty$ Suppose that $f(x), f'(x), f''(x)$ are continuous on $(0,+\infty)$ and $f(x) \geq \alpha > 0,$ and the improper integral $\displaystyle \int_{0}^{\infty} |f''(x)| dx$ exists. Proof that
    $$\displaystyle \int_0^\infty \dfrac{(f'(x))^2}{f(x)^{1980}} dx < \infty$$
My attempt is to use $\displaystyle \int \frac{f'(x)}{f(x)^{990}} dx = \frac{-1}{989 f(x)^{989}} $ but I got stuck. How can I use the continuity of $f , f', f''$  
 A: Updated answer (motivated/requested in the comments)
I'm sorry, but now the full solution is written "on the nose" of our (silent) OP.
The argument
Since the integrand $(f')^2/f^{1980}$ is non-negative, we only need to find a constant $C$ (not depending on $A$ or $B$) such that
$$
\int_A^B\frac{(f')^2}{f^{1980}}\,dx\leq C
$$
for all $0<A<1$ and $B>1$, since then we can conclude that the limits of $\int_A^B (f')^2/f^{1980}\,dx$ as $A\to 0^+$ and $B\to+\infty$ exists, and thus that $\int_0^{+\infty}(f')^2/f^{1980}\,dx$ is finite.
To have something with the second derivative, we integrate by parts,
$$
\int_A^{B}\frac{f'}{f^{1980}}f'\,dx=\Bigl[-\frac{1}{1979f^{1979}}f'\Bigl]_A^{B}+\int_A^{B}\frac{1}{1979f^{1979}}f''\,dx.
$$
We take care of the two terms separately.
The out-integrated term
We use the fundamental theorem of calculus and the triangle inequality to get
$$
|f'(B)|=\Bigl|f'(1)+\int_1^B f''(t)\,dt\Bigr|\leq |f'(1)|+\int_0^{\infty}|f''(t)|\,dt,
$$
and, likewise,
$$
|f'(A)|=\Bigl|f'(1)-\int_A^1 f''(t)\,dt\Bigr|\leq |f'(1)|+\int_0^{\infty}|f''(t)|\,dt,
$$
Thus, both $|f'(B)|$ and $|f'(A)|$ are bounded, independent of $A$ and $B$. Since $f(x)\geq \alpha>0$ it follows that the absolute value of the out-integrated term is bounded, say by a constant $C_1$, independent of $A$ and $B$.
The integral
Using the bound $f\geq \alpha>0$ together with $\int_0^{+\infty}|f''|\,dx<+\infty$ and the triangle inequality, we find that
$$
\begin{aligned}
\Bigl|\int_A^B\frac{1}{1979f^{1979}}f''\,dx\Bigr|
&\leq\frac{1}{1979\alpha^{1979}}\int_A^B|f''(x)|\,dx\\
&\leq\frac{1}{1979\alpha^{1979}}\int_0^{+\infty}|f''(x)|\,dx\\
&\leq C_2,
\end{aligned}
$$
where $C_2$ is independent of $A$ and $B$.
Conclusion
We conclude that we can take $C=C_1+C_2$.
A: Since $f > \alpha > 0 $ for all $x$ then $1/f^p < 1/\alpha^p$ for all $x$ and $p$. So
$$
\int_0^a (f')^2f^{-p}dx < \alpha^{-p}\int_0^a (f')^2dx, 
$$
for all $a > 0$. Evidently the claim is proven if the convergence of $\int |f''|dx$ is enough to conclude that $\int (f')^2dx$ converges. 
A: I am posting a partial answer:
We have $$\int_0^\infty \dfrac{(f'(x))^2}{f(x)^{1980}} dx.$$ We will integrate by parts: let $$u=f'(x)\text{ and }dv=f(x)^{-1980}f'(x).$$ Note that we have $$du=f''(x)\text{ and }v=-\frac{1}{1979}f(x)^{-1979}.$$ Thus, we have $$\int_0^\infty \dfrac{(f'(x))^2}{f(x)^{1980}} dx=\int_0^\infty udv=\lim_{x\to\infty}u(x)v(x)-u(0)v(0)-\int_0^\infty vdu.$$
We can check that $$\left|\int_0^\infty vdu\right|\le\int_0^\infty \left|-\frac{f''(x)}{1979}f(x)^{-1979}\right|dx\le\frac{1}{1979\alpha^{1979}}\int_0^\infty|f''(x)|dx\le\infty.$$
Thus, we have shown $$\int_0^\infty \dfrac{(f'(x))^2}{f(x)^{1980}} dx=\frac{f'(0)}{1979f(0)^{1979}}-\lim_{x\to\infty}\frac{f'(x)}{1979f(x)^{1979}}+\int_0^\infty\frac{f''(x)}{1979f(x)^{1979}}dx.$$
All that remains to be shown is that $\lim_{x\to\infty} u(x)v(x)$ exists. We can show $v(x)$ is bounded, as $$\liminf_{x\to\infty}v(x)=\liminf_{x\to\infty}-\frac{1}{1979}f(x)^{-1979}\ge\frac{-\alpha^{-1979}}{1979},$$ and $$\limsup_{x\to\infty}v(x)=0.$$ However, without $\lim_{x\to\infty}u(x)=0$, this still does not show that $\lim_{x\to\infty}u(x)v(x)$ exists. Perhaps additional assumptions are needed, or a clever trick.
EDIT: Per the observation of @mickep, since the integrand is non-negative, we need only bound the integral to show it converges. However, we have already shown this is bounded, up to showing $u(x)$ is bounded.
A: Since $f'(x) = f'(1)+\int_1^x f'',$ we see that $f'$ is bounded on $(0,\infty).$ It follows that
$$\int_0^1 \frac{(f')^2}{f^{1980}}<\infty.$$
On $[1,\infty)$ we can integrate by parts: For $b>1,$ we have
$$\int_1^b \frac{(f')^2}{f^{1980}} = \int_1^b f'f^{-1980}\cdot f' = \frac{f^{-1979}}{-1979}\cdot f'\ |_1^b + \int_1^b \frac{f^{-1979}}{1979}\cdot f''.$$
All we need to know is that the last expression is a bounded function of  $b\in [1,\infty).$ But this follows from $f\ge \alpha,$ the boundednes of $f',$ and the integrability of $f''.$
