the proof of an elementary problem in trignometry how do I prove the following?
$$\sin a_1 + \sin a_2 + \ldots + \sin a_n \lt n\sin\frac{k}{n}$$
where $k=\sum_{j=1}^n a_j$.
I can't do it by using the inequality of A.M. and G.M
err..here all the angles are ACUTE
 A: As JMoravitz noted, the inequality is not true. 
However, if one assumes that $a_1,\dots,a_n \in [0,\pi/2]$, an interval where $\sin (\cdot)$ is concave,  then it follows that 
$$\sin (\frac{1}{n} \sum_{j=1}^n a_j )\ge \frac 1n \sum_{j=1}^n \sin (a_j).$$ 
More generally, this is true when all (nonzero) $a_j$'s are on an interval where $\sin$ is concave.
Note however that if we take $a_1,\dots,a_n \in [-\pi/2,0]$, where $\sin$ is convex, the sign of the inequality above changes.  
A: Let $f''$ exist on an open interval containing $[a,b]$ and let $f''(x)\leq 0$ for $x\in [a,b].$ Let $x_1,...,x_n\in [a,b].$ Let $w_1,..,w_n>0$ with $\sum_1^nw_i=1.$ Then $\sum_1^nw_if(x_i)\leq f(\sum^n_0w_i x_i).$
Proof. Case $n=1$ is trivial. The case $x_1=x_i$ for all i from $1$ to $n$ is  also trivial. We prove  case $n=2$ with $x_1\ne x_2$ and then show that case $n>2$ can reduce to case $n=2$ by induction on $n$.
For $ n=2$  and $x_1< x_2$ the statement is equivalent to the assertion that  the graph of $f(x)$ for $x\in [x_1,x_2]$ has no point below the line segment joining $(x_1,f(x_1))$ to $(x_2,f(x_2)).$ 
By contradiction : Let $y=w_1x_1+w_2x_2$ and suppose $f(y)<w_1f(x_1)+w_2f_(x_2).$ $$ \text {Then }\quad  s_1<s_2 \text { where}$$ $$s_1=(f(y)-f(x_1))/(y-x_1),$$   $$s_2=(f(x_2)-f(y))/(x_2-y).$$ (Geometrically, $s_1$ is the slope of the line segment from $(x_1,f(x_1)$ to $(y,f(y))$ and $s_2$ is the slope of the line segment from $(y,f(y))$ to $(x_2,f(x_2).$)
 By the MVT there exists $z_1\in (x_1,y)$ and $z_2\in (y,x_2)$ with $f'(z_1)=s_1$ and $f'(z_2)=s_2.$ Applying the MVT to $f'$,there exists $z_3\in (z_1,z_2)$ with $$f''(z_3)=(f'(z_2)-f(z_1))/(z_2-z_1)=(s_2-s_1)/(z_2-z_1)>0,$$ contradicting the hypothesis that $f''\leq 0$ on $[a,b].$ 
For the case $n\geq 3,$ suppose it is true for case $n-1.$ 
Let $x^*_2=\sum^n_1w^*_i x_i$ where $w^*_i=w_i/\sum_2^n w_j$ for $2\leq i\leq n.$ By case $n-1$ we have $$\sum_2^n w^*_if(x_i)\leq f(\sum_2^nw^*_ix_i).$$
Let $x_2^*=\sum_2^nw_i^*x_1).$ We have $w_1x_1 +(1-w_1)x_2^*=$ $\sum_1^n w_ix_i.$
We have   $$\sum_1^n w_if(x_i)=w_1f(x_1)+(1-w_1)\sum_2^n w^*_if(x_i)\leq$$   $$\leq w_1 f(x_1)+(1-w_1) f(\sum^n_1 w_i^*x_i)=$$ $$= w_1f(x_1)+(1-w_1)f(x_2^*)\leq f(w_1 x_1+(1-w_1)x_2^*)=$$ $$=f(\sum_1^n w_i x_1). QED.$$
Although I proposed that $f''$ exists on an open interval containing  $[a,b],$ it is valid if $f$ is continuous on $[a,b]$, and $f''$ exists and is $\leq 0$ on $(a,b).$ For example if $[a.b]=[-1,1]$ and $f(x)=\sqrt {1-x^2}.$ This requires showing that the MVT applies for $f$ and for $f'$ when $x_1=a$ or $x_2=b,$ where $f'$ or $f''$ might not exist.
In the instance $w_i=1/n$ for $1\leq i\leq n$  we have $$\frac{1}{n} \sum_1^n f(x_i)\leq f(\frac {1}{n}\sum_1^nx_i).$$
A particular case $f(x)=\log x$ for $x>0$ yields the AGM inequality. Since $f''(x)=-1/x^2<0,$ we  have,for positive $x_1,...,x_n$ $$(\sum_1^n\log x_n)/n \leq \log (\sum_1^n x_n).$$ Exponentiating this, we have $$(\prod_1^n x_i)^{1/n}\leq (\sum_1^n x_i)/n.$$
