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I wonder how to prove rigorously the following conjecture without the use of Jensen's inequality:

By the use of the convexity/concavity of appropriate functions of 1 real variable, prove that:

  • $\forall x,y,z\in[0,\pi], \frac{sin(x)+\sin(y)+\sin(z)}{3} \le \sin(\frac{x+y +z}{3})$ (*)

While inequality in (*) becomes an inequality if and only if x=y=z.

I know that proof of this conjecture is straightforward if one uses Jensen's inequality for the function $f(x):=\sin(x)$ for $x\in[0,\pi]$ and knowing that function f(x) is at that interval concave.

Any help very appreciated!

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Hint: Write $(x+y+z)/3$ as ${1\over 3}x+{2\over 3}{y+z\over 2}$, and then apply the definition of concavity satisfied by the sine function on $[0,\pi]$ (namely $\sin(\lambda u+(1-\lambda)v)\ge\lambda\sin(u)+(1-\lambda)\sin(u)$ for $0<\lambda<1$ and $u,v\in[0,\pi]$) twice.

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