Explain how one can prove Bernoulli's inequality with the help of Mean Value Theorem?

Hint: The standard Mean Value Theorem for any continuous function: $ f(x) $ states that;
- If the function $ f(x) $ in the closed interval $ [a,b] $ is continuous function
- And if that function $ f(x) $ in the open interval $ (a,b) $ is a differentiable function
Then it should be understood that there exists a constant $ k $ in such a way that $ a < k < b $ and
 $ f'(k) = \dfrac{{f(b) - f(a)}}{{b - a}} $ or $ f(b) - f(a) = f'(k)(b - a) $
Then try using the function: $ f(x) = {(1 + x)^m} $ , also use $ f(x) - f(0) = x\;.f'(k) $

Complete step by step solution:
Let us note down everything we have and what we need;
- We know that the Bernoulli inequality states that;
 $ {(1 + x)^m} \geqslant 1 + mx\;\forall \;x \geqslant - 1,\;m \in N $
- We also know that for a continuous function the Mean Value Theorem says;
 $ f'(k) = \dfrac{{f(b) - f(a)}}{{b - a}} $ , for a function $ f(x) $ in the interval $ (a,b) $ where $ k $ is a constant.
- Now our aim is to prove the Bernoulli inequality with the help of Mean Value Theorem.
Let us relate the two, the Mean Value Theorem for the Bernoulli function is written in this way;
 $ f(x) = {(1 + x)^m} $ in the interval $ [0,x] $
 $ \Rightarrow \dfrac{{{{(1 + x)}^m} - 1}}{x} = m{(1 + k)^{m - 1}} $
Now we know that ;
 $ m{(1 + k)^{m - 1}} > m $
So we can write that;
 $ \Rightarrow \dfrac{{{{(1 + x)}^m} - 1}}{x} = m \times {(1 + k)^{m - 1}} > m $
 $ \Rightarrow {(1 + x)^m} - 1 > (m \times x) $
 $ \Rightarrow {(1 + x)^m} > 1 + mx $
So the proof is completed.
Now we can similarly prove for the case of $ - 1 \leqslant x < 0 $
Then for when $ x = 0 $ it can be proved easily.

Note: We should be clear about the Mean Value Theorem while solving questions related to it. Remember that, using this theorem we cannot find the actual value of $ k $ , but we will only be let known about the fact that there will be at least one number $ k $ that can satisfy the theorem. Also when we draw a tangent line to the function $ f(x) $ at $ x = k $ , we get the slope that is equal to $ f'(k) $ .