Prove that `2sinx + tanx ge3x`, for all `x in [0, pi/2]`.

Answer

To prove that \( 2\sin x + \tan x \geq 3x \) for all \( x \in [0, \frac{\pi}{2}] \), we can define a function and analyze its behavior.

### Step 1: Define the function
Let
\[ g(x) = 2\sin x + \tan x - 3x \]
We need to show that \( g(x) \geq 0 \) for all \( x \in [0, \frac{\pi}{2}] \).

### Step 2: Find the derivative
Now, we differentiate \( g(x) \):
\[ g'(x) = 2\cos x + \sec^2 x - 3 \]

### Step 3: Analyze the derivative
We need to analyze \( g'(x) \) to determine where \( g(x) \) is increasing or decreasing.

### Step 4: Simplify the derivative
We can rewrite \( g'(x) \):
\[ g'(x) = 2\cos x + \frac{1}{\cos^2 x} - 3 \]
To combine these terms, we can express everything in terms of \( \cos x \):
\[ g'(x) = \frac{2\cos^3 x + 1 - 3\cos^2 x}{\cos^2 x} \]
This gives us:
\[ g'(x) = \frac{2\cos^3 x - 3\cos^2 x + 1}{\cos^2 x} \]

### Step 5: Analyze the numerator
Let \( h(y) = 2y^3 - 3y^2 + 1 \) where \( y = \cos x \). We want to find the critical points of \( h(y) \) for \( y \in [0, 1] \).

### Step 6: Differentiate \( h(y) \)
Differentiating \( h(y) \):
\[ h'(y) = 6y^2 - 6y = 6y(y - 1) \]
Setting \( h'(y) = 0 \) gives us critical points at \( y = 0 \) and \( y = 1 \).

### Step 7: Evaluate \( h(y) \) at the endpoints
Now we evaluate \( h(y) \) at the endpoints:
- At \( y = 0 \):
\[ h(0) = 2(0)^3 - 3(0)^2 + 1 = 1 \]
- At \( y = 1 \):
\[ h(1) = 2(1)^3 - 3(1)^2 + 1 = 2 - 3 + 1 = 0 \]

### Step 8: Determine the behavior of \( h(y) \)
Since \( h'(y) \) changes sign from positive to negative, \( h(y) \) is increasing on \( [0, 1] \) and reaches a maximum at \( y = 0 \) and decreases to \( 0 \) at \( y = 1 \). Thus, \( h(y) \geq 0 \) for \( y \in [0, 1] \).

### Step 9: Conclusion
Since \( g'(x) \geq 0 \) for all \( x \in [0, \frac{\pi}{2}] \), it implies that \( g(x) \) is non-decreasing. Evaluating \( g(0) \):
\[ g(0) = 2\sin(0) + \tan(0) - 3(0) = 0 \]
Thus, \( g(x) \geq g(0) = 0 \) for all \( x \in [0, \frac{\pi}{2}] \).

Therefore, we conclude that:
\[ 2\sin x + \tan x \geq 3x \quad \text{for all } x \in [0, \frac{\pi}{2}]. \]