Let's solve these step by step using Lagrange's Mean Value Theorem or related reasoning.

(a) Prove that $\sin x \leq x$ for all $x \in [0, +\infty)$

Using Lagrange's Mean Value Theorem:

1. Define $f(x) = \sin x$. It is differentiable and continuous on $[0, x]$ for $x > 0$.
2. By the Mean Value Theorem, there exists some $c \in (0, x)$ such that: $f'(c) = \frac{f(x) - f(0)}{x - 0}$ Substituting $f(x) = \sin x$ and $f(0) = \sin(0) = 0$, we get: $\cos(c) = \frac{\sin x}{x}.$ Since $\cos(c) \leq 1$ for all $c \in [0, x]$, we deduce that: $\frac{\sin x}{x} \leq 1 \implies \sin x \leq x.$
3. This inequality holds for all $x > 0$. For $x = 0$, equality holds ($\sin 0 = 0$).

Thus, $\sin x \leq x$ for $x \in [0, +\infty)$.

(b) Prove that $(1 + x)^p \geq 1 + px$ for any $p \geq 1$ and $x \geq 0$

Using the Binomial Theorem:

1. Expand $(1 + x)^p$ using the Binomial Theorem: $(1 + x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \frac{p(p-1)(p-2)}{3!}x^3 + \dots$ Each term after $px$ is non-negative because $p \geq 1$ and $x \geq 0$. Hence: $(1 + x)^p \geq 1 + px.$

Alternatively, Using Lagrange's Mean Value Theorem:

1. Define $f(x) = (1 + x)^p - px - 1$. We need to show that $f(x) \geq 0$ for $x \geq 0$.
2. Compute $f'(x)$: $f'(x) = p(1 + x)^{p-1} - p.$ Since $(1 + x)^{p-1} \geq 1$ for $x \geq 0$, we have $f'(x) \geq 0$, so $f(x)$ is non-decreasing.
3. At $x = 0$, $f(0) = 0$. Since $f(x)$ is non-decreasing and $f(0) = 0$, $f(x) \geq 0$ for $x \geq 0$.

Thus, $(1 + x)^p \geq 1 + px$ for $p \geq 1$ and $x \geq 0$.

Would you like a detailed explanation of any of these steps or alternative proofs?

Related Questions:

1. How does Lagrange's Mean Value Theorem apply in other inequalities?
2. Can this inequality $(1 + x)^p \geq 1 + px$ be extended for $0 < p < 1$?
3. What is the geometric interpretation of $\sin x \leq x$?
4. How can Taylor expansions be used to prove these inequalities?
5. Are there other theorems that can be used to derive similar inequalities?

Tip:

When applying the Mean Value Theorem, always ensure that the function is continuous and differentiable on the relevant interval.