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.

Note that A,B,C are are positive acute angles, so both LHS and RHS are positive. Square both LHS and RHS, Then do LHS - RHS  we get.
$\cos2A + \cos2B + \cos2C + 2(\cos(A+B)+\cos(A+C)+\cos(B+C))$ note that $2(\cos(A+B)+\cos(A+C)+\cos(B+C))$ is always -ve for acute triangle (all these angles fall in second quadrant).
Check $\cos2A + \cos2B + \cos2C$, If all are greater than 45 deg. then this becomes negative. Suppose $A= 45 -\theta$ (in deg.), writing
$\cos2(45 -\theta) + \cos2(45 +\theta + a) + \cos2(45 +\theta + b)$ where $0\leq \theta < 45$ , $0< a + \theta < 45$, $0< b + \theta < 45$
This part becomes
$\sin2\theta - \sin(2a+2\theta) - \sin(2b+2\theta)$, all these angles fall in first quadrant, so this is also negative. So
$LHS < RHS$
