# In any triangle is $\sin A+\sin B+\sin C=\frac{3\sqrt3}{2}$ always

Well I came with an interesting proof. But I just want to verify it

From here we will get $$\sin A+\sin B+\sin C\leq \frac{3\sqrt3}{2}.$$

and from this I get $$\sin A+\sin B+\sin C\geq \frac{3\sqrt3}{2}.$$

Now the equation to be satisfied, only equality condition should hold.

So in an acute angled triangle $$\sin A+\sin B+\sin C=\frac{3\sqrt3}{2}.$$

Is there any fallacy in this convention.

• Take half a square. Then $$\sin A + \sin B + \sin C = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \neq \frac{3\sqrt{3}}{2}$$ – Theo Bendit Feb 15 at 7:58
• To echo Theo's comment, don't start looking for a general proof before you've tried a few specific examples. – Austin Mohr Mar 14 at 21:34

There's an implied claim in the second block, that $$\sqrt[3]{\sin A\sin B\sin C}\ge \frac{\sqrt{3}}{2}$$. That claim is false. In fact, $$\sqrt[3]{\sin A\sin B\sin C}\le \frac{\sqrt{3}}{2}$$ with equality only when $$A=B=C=60^\circ$$. In an acute triangle, that quantity can get arbitrarily close to zero - consider a triangle with angles $$\epsilon, 90^\circ-\frac{\epsilon}{2}, 90^\circ-\frac{\epsilon}{2}$$. The product of sines in that triangle is less than $$\sin\epsilon$$, which goes to zero as $$\epsilon\to 0$$.
I take it (from the last inequality claimed in the second block) you were asked to prove that $$\sin A+\sin B+\sin C > 2$$ in an acute triangle? That's true, and it's as strong as we can possibly have. Equality there is approached by the triangle I mentioned, in which the sines approach $$0,1,1$$.
In an acute angled triangle, it should be noted that $$\sqrt[3]{ sinA sinB sinC} \geq \ {\sqrt{3}\over2}$$ implies that the angles are $$\geq 60$$. In this case, the sum of the angles are going to be $$\geq 180$$. So, such a triangle will not exist (except for the case of an equilateral triangle, where your proposition seems right).