# In a triangle, prove that $\sin A + \sin B + \sin C \leq 3 \sin \left(\frac{A+B+C}{3}\right)$

Prove that for any $$\Delta ABC$$ we have the following inequality:

$$\sin A + \sin B + \sin C \le 3 \sin \left(\frac{A+B+C}{3}\right)$$

Could you use AM-GM to prove that?

• Try some simple triangles. Is it true? May 13, 2019 at 11:39
• Use Jensen's inequality May 13, 2019 at 11:41
• Note that $(A+B+C)/3= 60^\circ$, so the right hand side is equal to $3\sqrt 3/2$. It is well-known that $\sin A+\sin B+\sin C\leqslant 3\sqrt 3/2$, so it should be $\leqslant$ insted of $\geqslant$ in the question.
– SMM
May 13, 2019 at 11:52
• @DragunityMAX why May 13, 2019 at 12:00
• which triangle inscribed in a fixed circle have the largest perimeter ? May 13, 2019 at 12:28

I'm sure there's a different way to approach this question, but here's one way using the graph of $$\sin x$$: Consider $$3$$ points on the graph for $$x\in(0,\pi)$$. They are $$(A,\sin A)$$, $$(B,\sin B)$$, and $$(C,\sin C)$$ as shown. $$A,B,C$$ are such that $$A+B+C=\pi$$.

(Ignore the fact that $$A,B$$ and $$C$$ are angles of a triangle, they are just some values of $$x$$ that satisfy the condition $$A+B+C=\pi$$)

$$A, B$$ and $$C$$ are plotted on the graph of $$y=\sin x$$ and are joined to form a triangle, as shown.

Consider the centroid of the triangle, $$G$$, given by: $$G=\left(\dfrac{A+B+C}{3}, \dfrac{\sin A+\sin B+\sin C}{3}\right)$$ Draw a line $$PG$$ as shown at $$x=\dfrac{A+B+C}{3}$$ (or $$\dfrac{\pi}{3}$$). This line intersects the curve at the point $$P$$ given by: $$P=\left(\dfrac{A+B+C}{3}, \sin \left(\dfrac {A+B+C}{3}\right)\right)$$

From the figure, it is evident that the $$y$$-value of $$P$$ $$>$$ the $$y$$-value of $$G$$. So we obtain the inequality $$\sin \left(\dfrac {A+B+C}{3}\right) >\dfrac{\sin A+\sin B+\sin C}{3}$$

Note that, for the case where $$A=B=C$$, we can deduce that $$A=B=C= \dfrac{\pi}{3}$$ and therefore $$A,B$$ and $$C$$ will coincide with point $$P$$ on the curve. For this particular case, we can deduce that $$\sin \left(\dfrac {A+B+C}{3}\right) =\dfrac{\sin A+\sin B+\sin C}{3}$$

Combining both the inequalities obtained, we get the desired result: $$\sin \left(\dfrac {A+B+C}{3}\right) \geq \dfrac{\sin A+\sin B+\sin C}{3}$$ or $$3 \sin \left(\dfrac {A+B+C}{3}\right) \geq \sin A+\sin B+\sin C$$

Credit for the answer: "Play with Graphs" by Amit Agarwal.

• it's awesome mate ^ ^, thanks for your explaining ! May 14, 2019 at 13:22
• no problem, glad to help! May 14, 2019 at 14:56

We know in a triangle$$ABC, 0 \lt A,B,C \lt \pi$$. So $$sin A , sinB, sinC \gt 0$$. Moreover in $$[0,\pi],$$ so $$(sinx)''=-sinx \lt 0$$, which is concave downwards. Using Jensen's inequality for concave downwards, we have:$$sin(\frac {A+B+C} 3)\ge \frac {sinA+sinB+sinC} 3$$ or $$sinA +sinB+sinC \le 3\sqrt 3 /2$$