IIT-JEE 1997Question 1
Let $0<A_i<\pi$ for $i=1,2,\ldots,n$. Use mathematical induction to prove that $$\sin A_1+\sin A_2+\cdots+\sin A_n\le n\sin\left(\dfrac{A_1+A_2+\cdots+A_n}{n}\right)$$ where $n\ge1$ is a natural number. {You may use the fact that $p\sin x+(1-p)\sin y\le\sin[px+(1-p)y]$, where $0\le p\le1$ and $0\le x,y\le\pi$.}
Trigonometric Ratio & Identities
Answerproved
Solution
Concept. Induction on $n$. The step from $n$ to $n+1$ uses the given inequality with weight $p=\frac{n}{n+1}$. Formulas used. (1) Given: $p\sin x+(1-p)\sin y\le\sin\left[px+(1-p)y\right]$ for $0\le p\le1$, $0\le x,y\le\pi$Step 1 - Base case. For $n=1$ both sides are $\sin A_1$. Step 2 - Hypothesis. Assume, for some $n$, that $\sum_{i=1}^{n}\sin A_i\le n\sin\frac Sn$, where $S=A_1+\cdots+A_n$. Step 3 - Step to $n+1$. Put $x=\frac Sn\in(0,\pi)$, $y=A_{n+1}\in(0,\pi)$, $p=\frac{n}{n+1}$. Then $$\sum_{i=1}^{n+1}\sin A_i\le n\sin x+\sin y=(n+1)\left[p\sin x+(1-p)\sin y\right].$$ Step 4 - Use (1). $px+(1-p)y=\dfrac{S+A_{n+1}}{n+1}$, so $$\sum_{i=1}^{n+1}\sin A_i\le(n+1)\sin\frac{A_1+\cdots+A_{n+1}}{n+1}.$$ This is the statement for $n+1$; by induction it holds for every $n\ge1$. Hence proved.