$f(c)\int_{1}^{\infty} f(x) \phi(x) dx =g(c)\int_{1}^{\infty} g(x) \phi(x) dx$ 
If $f$ and $g$ are continuous on $[1,\infty)$ and $\phi(x)\geq0$ and be integrable in $[1,\infty) $. Also $ \int_{1}^{\infty}\phi(x) dx \neq 0$ and $ \int_{1}^{\infty} g(x) \phi(x) dx \neq 0$ then there exists some
$c\in(1,\infty)$ such that
$$
g(c)\int_1^\infty f(x) \phi(x) dx =
     f(c)\int_1^\infty g(x) \phi(x) dx
$$

MY TRY - Let
$$
\frac{\int_1^\infty f(x) \phi(x) dx}
     {\int_1^\infty g(x) \phi(x) dx} = \lambda
$$
so
$$
\int_{1}^{\infty}  (f(x)-\lambda g(x)) \phi(x)dx =0
$$
After this i am stuck.
 A: If
$$\int_1^\infty (f(x) - \lambda g(x))\phi(x)dx =0$$
With $\phi \geq 0$ (and $\phi \not\equiv 0$), then by continuity, there must be some $c\in(1,\infty)$ such that $f(c) - \lambda g(c) = 0$, otherwise if $f(x) -\lambda g(x) >0$ for all $x\in(1,\infty)$, the integral would be positive. If $f(x) - \lambda g(x) < 0$ for all $x\in(1,\infty)$ the integral would be negative.
Edit:
I will discuss the case when $g(c) =0$ as this implies $f(c)=0$ and $\lambda =1$. I have shown
$$\exists c\in (1,\infty), \text{ such that } g(c)\int_1^\infty f(x)\phi(x)dx = f(c)\int_1^\infty g(x)\phi(x)dx, \quad (1)$$
As a counter example to the original claim
$$\exists c\in (1,\infty), \text{ such that } \frac{\int_1^\infty f(x)\phi(x)dx}{\int_1^\infty g(x) \phi(x)dx} = \frac{f(c)}{g(c)}, \quad (2)$$
Consider taking $\phi = 1_{[1,4]}$, and $f$ to be a bump function with support in $[1,2]$ and $g$ a shifted bump function of $f$ with support in $[3,4]$. This implies $\lambda =1$ and
\begin{align}
\int_1^\infty(f(x) - \lambda g(x))\phi(x)dx &= \int_1^4(f(x) - g(x))dx 
\\
&= \int_1^2 f(x)dx - \int_3^4g(x)dx =0
\end{align}
But there is no point $c\in [1,\infty)$ such that
$$\frac{f(c)}{g(c)} = 1$$
since $f \equiv 0$ on the support of $g$. Therefor you cannot guarantee (2), but you Can guarentee the exitence of $c\in (1,\infty)$ for the equality (1). In this example $f(2.5) = g(2.5) = 0$.
