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

$$\begin{array}{l} {\text { Applied at function } y=\sin x,} \\ {\text { We have, } \sin \left(\frac{A+B+C}{3}\right) \geq \frac{\sin A+\sin B+\sin C}{3}} \end{array}$$

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

$$\begin{array}{l} {\text { Also by A.M. } \geq \text { G.M. in an acute angled triangle }} \\ {\frac{\sin A+\sin B+\sin C}{3} \geq \sqrt[3]{\sin A \sin B \sin C}} \\ {\Rightarrow \sin A+\sin B+\sin C \geq 3(\sqrt[3]{\sin A \sin B \sin C})} \\ {\Rightarrow \sin A+\sin B+\sin C \geq 3\left(\frac{\sqrt{3}}{2}\right)=\frac{3 \sqrt{3}}{2}>2} \end{array}$$

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.

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    $\begingroup$ Take half a square. Then $$\sin A + \sin B + \sin C = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \neq \frac{3\sqrt{3}}{2}$$ $\endgroup$ Commented Feb 15, 2019 at 7:58
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    $\begingroup$ To echo Theo's comment, don't start looking for a general proof before you've tried a few specific examples. $\endgroup$ Commented Mar 14, 2019 at 21:34

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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 $\epsilon\to 0$.

Naturally, the arguments that follow from that don't work.

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 $0,1,1$.

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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).

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