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?

  • $\begingroup$ Try some simple triangles. Is it true? $\endgroup$
    – Empy2
    Commented May 13, 2019 at 11:39
  • 1
    $\begingroup$ Use Jensen's inequality $\endgroup$
    – user610336
    Commented May 13, 2019 at 11:41
  • 2
    $\begingroup$ 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. $\endgroup$
    – SMM
    Commented May 13, 2019 at 11:52
  • $\begingroup$ @DragunityMAX why $\endgroup$ Commented May 13, 2019 at 12:00
  • $\begingroup$ which triangle inscribed in a fixed circle have the largest perimeter ? $\endgroup$
    – HK Lee
    Commented May 13, 2019 at 12:28

2 Answers 2


I'm sure there's a different way to approach this question, but here's one way using the graph of $\sin x $:

A graph of y = sin x with a triangle on it

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.

  • $\begingroup$ it's awesome mate ^ ^, thanks for your explaining ! $\endgroup$ Commented May 14, 2019 at 13:22
  • $\begingroup$ no problem, glad to help! $\endgroup$ Commented 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 $sinA +sinB+sinC \le 3\sqrt 3 /2$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .