If $f(x)$ is positive and $f'(x)<0$, $f"(x)>0$ for all $x>0$, prove that $\frac{1}{2}f(1)+\int_1^nf(x)dx<\sum_{r=1}^nf(r)$ 
If $f(x)$ is positive and $f'(x)<0$, $f"(x)>0$ for all $x>0$, prove that $$\frac{1}{2}f(1)+\int_1^nf(x)dx<\sum_{r=1}^nf(r)<f(1)+\int_1^nf(x)dx$$

Since function is decreasing, then area of rectangles $$f(2)[2-1]+f(3)[3-2]+...<\int_1^nf(x)dx$$
Hence, $$\sum_{r=2}^nf(r)<\int_1^nf(x)dx$$
Adding $f(1)$ on both sides, 
$$\sum_{r=1}^nf(r)<f(1)+\int_1^nf(x)dx$$
But how do I get the second inequality : $$\frac{1}{2}f(1)+\int_1^nf(x)dx<\sum_{r=1}^nf(r)$$

$f'' > 0$ means that $f$ is strictly convex, so the graph of $f$ restricted to each interval $[r, r+1]$ lies below the straight line connecting 
$(r, f(r))$ and $(r+1, f(r+1))$:
$$
f(x) \le (r+1-x) \, f(r) + (x-r) \, f(r+1)
$$
with equality only at the endpoints of the interval.
The integral of a linear function over an interval is just
the value at the midpoint, multiplied by the length of the interval.
It follows that
$$
\int_r^{r+1}f(x)dx < \frac 12 \bigl( f(r) + f(r+1)\bigr ) 
$$
and therefore
$$
\int_1^nf(x)dx < \frac 12 f(1) + f(2) + \dots + f(n-1) + \frac 12 f(n) 
$$
or 
$$
\frac 12 f(1) + \int_1^nf(x)dx + \frac 12 f(n) < \sum_{r=1}^n f(r) \, .
$$
The conclusion now follows since $\frac 12 f(n) > 0$.
