Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$ 
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
  $$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$

To begin, I proved a lemma: 

Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.

Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.

How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?

Any help or hint will be appreciated.

For a nice algebraic way to prove it, consider the inequality
$$\sin x\geq \frac{2x}{\pi}, x\in[0,\frac{\pi}{2}]$$
and apply for $A,B,C>\frac{\pi}{4}$:
$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq\frac{2}{\pi}(A+B+C-\frac{3\pi}{4})=\frac{1}{2}>0$$
When we assume wlog $A<\frac{\pi}{4}$, we can instead assert that:
$$\sin(A-\frac{\pi}{4})\geq A-\frac{\pi}{4}$$
and then
$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq 1-\frac{\pi}{4}+(1-\frac{2}{\pi})A>1-\frac{\pi}{4}>0$$
This is not the sharpest bound.  Actually one can show that the minimum value this expression is achieved for A=B=C and therefore
$$\require{cancel}\xcancel{\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq 3\sin(\frac{\pi}{12})=\frac{3}{2}\sqrt{2-\sqrt{3}}}$$
EDIT: After a while, I finally noticed that, while the solution to the problem satisfactorily addresses the question asked, a mistake above has obscured a clear elementary solution to finding maxima and minima to this function. We want to show that whenever $A,B,C>0$ with $A+B+C=\pi$,
$$\frac{\sqrt{2}}{2}<\Delta=\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\leq 3\sin\frac{\pi}{12}$$

In the figure-where the contours of $\Delta(A,B)$ are depicted- one can clearly see that, within the area of interest denoted by a red triangle, the function attains a maximum. The minima are attained on the red triangle itself. As of now I haven't found a satisfactory algebraic/elementary approach to obtain the minima and maxima in the entire triangle interior and boundary (but with calculus it is pretty easy to demonstrate).
To get the maximum in a part of the interior, assume that $A,B,C>\frac{\pi}{4}$. In this case we can apply Jensen's inequality for the concave function $f(x)=\sin(x)~,~ x\in (0,\pi)$ and we obtain
$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\leq 3\sin(\frac{A+B+C}{3}-\frac{\pi}{4})=3\sin\frac{\pi}{12}$$.
