<s>[INST] 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.
 [/INST] A purely "trig-bashing" algebraic way:
We have
\begin{align*}
&\sin\left(B-\frac\pi4\right)+\sin\left(C-\frac\pi4\right)\\
&=2\sin\left(\frac{B+C}2-\frac\pi4\right)\cos\frac{B-C}2\\
&=2\sin\left(\frac\pi4-\frac{A}2\right)\cos\frac{B-C}2.
\end{align*}
So we want to prove
$$
2\cos\frac{B-C}2-1>0
$$
when $A<\pi/4$, $B,C$ acute.
Note that we have $\lvert B-C\rvert$ is at most $A$ (only in the degenerate case when $B$ or $C$ is $\pi/2$), so
$$
2\cos\frac{B-C}2-1>2\cos\frac{A}2-1>2\cos\frac\pi8-1>0
$$
as desired.
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