The Ultimate Monotonicity Trick: Proving Inequalities using Derivatives
Master this powerful technique that solves 80% of JEE inequality problems. Transform complex inequality proofs into simple derivative checks.
Why This Technique is a Game-Changer
Traditional inequality proofs often involve clever algebraic manipulations that are hard to spot. The monotonicity approach provides a systematic, guaranteed method that works for a wide range of problems.
🎯 JEE Advantage
- Solves problems that seem to require "magical" algebraic tricks
- Provides a step-by-step mechanical approach
- Works for both strict and non-strict inequalities
- Can prove multiple inequalities simultaneously
- Especially powerful for trigonometric and exponential inequalities
🚀 Technique Navigation
The Monotonicity Principle
📚 Mathematical Foundation
Theorem: If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then:
• $f'(x) > 0$ on $(a,b) \Rightarrow f$ is strictly increasing on $[a,b]$
• $f'(x) < 0$ on $(a,b) \Rightarrow f$ is strictly decreasing on $[a,b]$
💡 How This Proves Inequalities
To prove $f(x) \geq g(x)$ for $x \in [a,b]$:
Step 1: Define $h(x) = f(x) - g(x)$
Step 2: Show $h'(x) \geq 0$ (or $\leq 0$) on $(a,b)$
Step 3: Check $h(a) \geq 0$ (or $h(b) \geq 0$)
Step 4: Conclude $h(x) \geq 0$ for all $x \in [a,b]$
🎯 Key Insight
Instead of directly comparing $f(x)$ and $g(x)$, we study the monotonic behavior of their difference. If the difference starts non-negative and only increases, it must remain non-negative!
The 4-Step Monotonicity Method
Step 1: Reformulate the Inequality
Express the inequality as $h(x) \geq 0$ or $h(x) \leq 0$
Example: To prove $\sin x \geq x - \frac{x^3}{6}$ for $x \geq 0$
Define $h(x) = \sin x - x + \frac{x^3}{6}$
Step 2: Compute the Derivative
Find $h'(x)$ and simplify as much as possible
Example: $h'(x) = \cos x - 1 + \frac{x^2}{2}$
Step 3: Analyze the Derivative
Determine the sign of $h'(x)$ on the interval
Example: $h''(x) = -\sin x + x \geq 0$ for $x \geq 0$
So $h'(x)$ is increasing, and $h'(0) = 0$, thus $h'(x) \geq 0$
Step 4: Use Initial/Boundary Values
Check $h(x)$ at the starting point and conclude
Example: $h(0) = 0$ and $h'(x) \geq 0$, so $h(x) \geq 0$ for $x \geq 0$
Example 1: Exponential Inequality
Prove that $e^x \geq 1 + x$ for all $x \in \mathbb{R}$
Proof using Monotonicity
Step 1: Define $f(x) = e^x - 1 - x$
We need to show $f(x) \geq 0$ for all $x \in \mathbb{R}$
Step 2: Compute derivative: $f'(x) = e^x - 1$
Step 3: Analyze $f'(x)$:
• $f'(x) = 0$ when $x = 0$
• $f'(x) < 0$ for $x < 0$
• $f'(x) > 0$ for $x > 0$
Step 4: Find minimum:
$f'(x)$ changes from negative to positive at $x = 0$
So $x = 0$ is a minimum point
$f(0) = e^0 - 1 - 0 = 0$
Step 5: Conclusion:
Since $f(0) = 0$ is the minimum value, $f(x) \geq 0$ for all $x$
Therefore, $e^x \geq 1 + x$ for all $x \in \mathbb{R}$ ✓
Example 2: Trigonometric Inequality
Prove that $\frac{2x}{\pi} \leq \sin x \leq x$ for $0 \leq x \leq \frac{\pi}{2}$
Proof using Monotonicity
Part 1: Prove $\sin x \leq x$
Step 1: Define $f(x) = x - \sin x$
Step 2: $f'(x) = 1 - \cos x \geq 0$ for all $x$
Step 3: $f(0) = 0$, and $f$ is increasing
Step 4: So $f(x) \geq 0 \Rightarrow x \geq \sin x$ ✓
Part 2: Prove $\frac{2x}{\pi} \leq \sin x$
Step 1: Define $g(x) = \sin x - \frac{2x}{\pi}$
Step 2: $g'(x) = \cos x - \frac{2}{\pi}$
Step 3: $g'(x) = 0$ when $\cos x = \frac{2}{\pi} \Rightarrow x = \cos^{-1}(\frac{2}{\pi})$
Step 4: Check endpoints:
$g(0) = 0$, $g(\frac{\pi}{2}) = 1 - 1 = 0$
Step 5: Since $g(x)$ is concave down, minimum at endpoints
Step 6: So $g(x) \geq 0 \Rightarrow \sin x \geq \frac{2x}{\pi}$ ✓
Example 3: Logarithmic Inequality
Prove that $\frac{x}{1+x} \leq \ln(1+x) \leq x$ for $x > -1, x \neq 0$
Proof using Monotonicity
Part 1: Prove $\ln(1+x) \leq x$
Step 1: Define $f(x) = x - \ln(1+x)$
Step 2: $f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x}$
Step 3: Sign analysis:
For $x > 0$: $f'(x) > 0 \Rightarrow f$ increasing
For $-1 < x < 0$: $f'(x) < 0 \Rightarrow f$ decreasing
Step 4: $f(0) = 0$, so $f(x) \geq 0$ for all $x > -1$
Step 5: Thus $\ln(1+x) \leq x$ ✓
Part 2: Prove $\frac{x}{1+x} \leq \ln(1+x)$
Step 1: Define $g(x) = \ln(1+x) - \frac{x}{1+x}$
Step 2: $g'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2}$
Step 3: Sign analysis:
For $x > 0$: $g'(x) > 0 \Rightarrow g$ increasing
For $-1 < x < 0$: $g'(x) < 0 \Rightarrow g$ decreasing
Step 4: $g(0) = 0$, so $g(x) \geq 0$ for all $x > -1$
Step 5: Thus $\frac{x}{1+x} \leq \ln(1+x)$ ✓
🎯 Key JEE Applications
Direct Applications:
- Inequality proofs in function theory
- Range determination for complex functions
- Limit evaluation using squeeze theorem
- Series convergence tests
- Optimization problems with constraints
Indirect Applications:
- Functional equations analysis
- Differential equations behavior study
- Curve sketching and analysis
- Approximation theory error bounds
- Physics applications in motion analysis
📝 Technique Mastery Checklist
Check which aspects you've mastered:
Note: If you checked all 6, you've mastered this powerful technique!
🚀 Advanced Tips & Tricks
When First Derivative Fails:
- Try second derivative test for concavity
- Use higher order derivatives if needed
- Consider Taylor series expansions
- Look for symmetry or periodicity
- Try substitution techniques
Time-Saving Strategies:
- Memorize common inequalities and their proofs
- Practice mental derivative calculation
- Develop pattern recognition for function types
- Use graphical intuition to guide algebraic work
- Learn quick sign analysis methods
🎯 Test Your Understanding
Try these problems using the monotonicity technique:
1. Prove that $\tan x \geq x$ for $0 \leq x < \frac{\pi}{2}$
2. Prove that $x - \frac{x^2}{2} \leq \ln(1+x) \leq x$ for $x \geq 0$
3. Show that $\frac{\sin x}{x}$ is decreasing on $(0, \pi)$
Master This Power Technique!
The monotonicity method is your secret weapon for inequality problems in JEE Advanced