Let 0

Q: Wait, why can't we just use AM-GM inequality here?

That is a fantastic question! AM-GM relates the arithmetic mean and geometric mean of non-negative numbers. However, our problem involves the sine function, which is not a simple product. We are dealing with the sum of sines, and the sine function is concave, not logarithmic or exponential. This is why we need the power of Jensen's Inequality, which is specifically designed for convex or concave functions.

Q: Why is the sine function concave on $(0, \pi)$?

Think about the second derivative! The derivative of $\sin x$ is $\cos x$, and the second derivative is $-\sin x$. On the interval $(0, \pi)$, $\sin x$ is always positive, so $-\sin x$ is always negative. A negative second derivative means the function is strictly concave—it bends downwards, like a bridge or an arch. This geometric property is exactly what allows us to prove the inequality.

Q: How do I know when to use induction versus Jensen's Inequality directly?

Both are valid! Induction is a powerful, step-by-step way to build the proof from the ground up, which is excellent for understanding the mechanics. Jensen's Inequality is a more direct, elegant tool that generalizes this result for any concave function. In a JEE context, if the problem explicitly asks for induction, you must follow that path, but knowing Jensen's Inequality gives you the 'big picture' intuition that makes the induction steps feel natural.

Question

Let $0 < A_i < \pi$ for $i = 1, 2, ...., n$. Use mathematical induction to prove that $\sin A_1 + \sin A_2 + .... + \sin A_n \le n \sin \frac{A_1 + A_2 + .... + A_n}{n}$ where $n \ge 1$ is a natural number. {You may use the fact that $p \sin x + (1 - p) \sin y \le \sin [px + (1 - p)y]$, where $0 \le p \le 1$ and $0 \le x, y \le \pi$}.

Sigma Percentile · Accepted Answer

### The Geometry of the Sine Arch: A Journey into Concavity

My dear student, welcome to a beautiful exploration of one of the most elegant inequalities in trigonometry. Today, we are not just solving a problem; we are uncovering the hidden geometry of the sine function.

We are tasked with proving that for any angles $0 < A_i < \pi$, the sum of their sines is bounded by $n$ times the sine of their average:

$$ \sum_{i=1}^n \sin A_i \le n \sin \left( \frac{\sum_{i=1}^n A_i}{n} \right) $$

This is a classic, and it is the gateway to understanding **Jensen's Inequality**.

### Phase 1: Visualizing the Playground

Imagine you are standing on a coordinate plane, looking at the graph of $y = \sin x$ between $0$ and $\pi$. You see a beautiful, symmetric arch.

If you were to place a ruler between any two points on this arch, the ruler would lie entirely below the curve. This is the definition of a **strictly concave function**.

In the language of calculus, because the second derivative:

$$ \frac{d^2}{dx^2}(\sin x) = -\sin x $$

is negative on this interval, the curve is always 'bulging' upwards. This geometric reality is our most powerful weapon.

### Phase 2: The Inductive Leap

We begin our proof with the base case, $n=1$. The inequality becomes $\sin A_1 \le 1 \cdot \sin (A_1/1)$, which is simply $\sin A_1 \le \sin A_1$. It is trivially true, a solid foundation upon which we can build.

Now, we assume the statement holds for some natural number $k$. We define the average of the first $k$ angles as $\bar{A}_k = \frac{\sum_{i=1}^k A_i}{k}$.

Our inductive hypothesis is:

$$ \sum_{i=1}^k \sin A_i \le k \sin \bar{A}_k $$

This is our 'known' truth.

### Phase 3: The Weighted Average Trick

Now, we step up to $n=k+1$. We write the sum as:

$$ \sum_{i=1}^{k+1} \sin A_i = \left( \sum_{i=1}^k \sin A_i \right) + \sin A_{k+1} $$

Using our hypothesis, we bound the first part:

$$ \sum_{i=1}^{k+1} \sin A_i \le k \sin \bar{A}_k + \sin A_{k+1} $$

Here is where the magic happens. We have $k \sin \bar{A}_k + \sin A_{k+1}$, but we need to use the concavity property, which requires a weighted average. We multiply and divide by $(k+1)$ to get:

$$ (k+1) \left[ \frac{k}{k+1} \sin \bar{A}_k + \frac{1}{k+1} \sin A_{k+1} \right] $$

Now, look at the coefficients: $p = \frac{k}{k+1}$ and $(1-p) = \frac{1}{k+1}$. They sum to exactly $1$! We can now apply the concavity property:

$$ p \sin x + (1-p) \sin y \le \sin(px + (1-p)y) $$

### Phase 4: The Final Destination

By applying this property, our expression becomes:

$$ (k+1) \sin \left( \frac{k}{k+1} \bar{A}_k + \frac{1}{k+1} A_{k+1} \right) $$

Look closely at the argument inside the sine function. The expression $\frac{k \bar{A}_k + A_{k+1}}{k+1}$ is simply the arithmetic mean of all $k+1$ angles, which we can call $\bar{A}_{k+1}$.

Thus, we have shown that:

$$ \sum_{i=1}^{k+1} \sin A_i \le (k+1) \sin \bar{A}_{k+1} $$

The induction is complete! We have traversed the path from a simple base case to a general truth, using the elegant geometry of the sine curve. Remember, mathematics is not just about symbols; it is about seeing the shapes behind the equations.

Visualized Solution

Understanding the Domain

The Hint: Concavity Property

Weighted Averages

Base Case: $n = 1$

Inductive Hypothesis

Inductive Step: $n = k + 1$

Formatting to Match the Hint

Mapping to the Hint

Applying the Inequality

Simplifying the Argument

Conclusion of the Proof

The Sigma Insight: Maxima and Minima

The Geometry of the Sine Arch

Phase 1

Phase 2

Phase 3

Phase 4

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