Let `0 lt A_(i) pi` for `i = 1,2,"……"n`. Use mathematical induction to prove that `sin A_(1) + sin A_(2)+ "….." + sin A_(n) le n sin ((A_(1) + A_(2) +

For `n = 1`, the inequality becomes
`sinA_(1) le sin A_(1)`, which is clearly true.
Let us assume that the inequility holds for `n = k` where k is some positive integer.
Then `sinA_(1) + sinA_(2) + "……"+sinA_(n) le k sin ((A_(1) + A_(2) +"....." + A_(k))/(k))"......"(1)`
Adding `sin A_(k+1)` on both sides, we get
`sin A_(1) + sin A_(2) + "......" + sin A_(k) + suin A_(k+1) le k sin((A_(1) + A_(2) +"......."+A_(k))/(k)) + sin A_(k+1)`
Now, `k sin ((A_(1) + A_(2) + "......." + A_(k))/(k)) + sin A_(k+1)`
`= (k+1)[(k)/(k+1)sin alpha+(1)/(k+1) sin A_(k+1)]`,
where `alpha = (A_(1) + A_(2) + "......" + A_(g))/(k)`
` le (k + 1 )sin{(1+(1)/(k+1)) alpha + (1)/(k+1) A_(k+1)}`
{Using `p sin x + (1-p) sin y le sin[px + (1-p)y]}`
`= (k+1)sin((A_(1)+A_(2)+"......."A_(k+1))/(k+1))`
Thus, the inequality hold for `n = k +1`.
Hence, by the principle of mathematical induction, the inequality holds for all `n in N`.