The inequality `2^(sintheta) + 2 ^(cos theta ) ge 2^(1- ( 1//sqrt(2)))` holds for all real values of `theta`

Answer

To solve the inequality \( 2^{\sin \theta} + 2^{\cos \theta} \geq 2^{1 - \frac{1}{\sqrt{2}}} \) for all real values of \( \theta \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Apply the AM-GM Inequality**:
According to the AM-GM inequality, for any non-negative numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
In our case, let \( a = 2^{\sin \theta} \) and \( b = 2^{\cos \theta} \). Thus, we have:
\[ \frac{2^{\sin \theta} + 2^{\cos \theta}}{2} \geq \sqrt{2^{\sin \theta} \cdot 2^{\cos \theta}} \]

2. **Simplify the Right Side**:
The right-hand side can be simplified:
\[ \sqrt{2^{\sin \theta} \cdot 2^{\cos \theta}} = \sqrt{2^{\sin \theta + \cos \theta}} = 2^{\frac{\sin \theta + \cos \theta}{2}} \]
Therefore, we can rewrite the inequality as:
\[ 2^{\sin \theta} + 2^{\cos \theta} \geq 2 \cdot 2^{\frac{\sin \theta + \cos \theta}{2}} = 2^{1 + \frac{\sin \theta + \cos \theta}{2}} \]

3. **Estimate the Maximum Value of \( \sin \theta + \cos \theta \)**:
The maximum value of \( \sin \theta + \cos \theta \) occurs when \( \theta = \frac{\pi}{4} \) or \( \theta = \frac{5\pi}{4} \), giving:
\[ \sin \theta + \cos \theta \leq \sqrt{2} \]
Therefore, we have:
\[ \frac{\sin \theta + \cos \theta}{2} \leq \frac{\sqrt{2}}{2} \]

4. **Substitute the Maximum Value**:
Substituting this maximum value into our inequality gives:
\[ 2^{\sin \theta} + 2^{\cos \theta} \geq 2^{1 + \frac{\sqrt{2}}{2}} \]

5. **Compare with the Right-Hand Side**:
The right-hand side of our original inequality is \( 2^{1 - \frac{1}{\sqrt{2}}} \). We need to show:
\[ 2^{1 + \frac{\sqrt{2}}{2}} \geq 2^{1 - \frac{1}{\sqrt{2}}} \]
This simplifies to:
\[ 1 + \frac{\sqrt{2}}{2} \geq 1 - \frac{1}{\sqrt{2}} \]
Rearranging gives:
\[ \frac{\sqrt{2}}{2} + \frac{1}{\sqrt{2}} \geq 0 \]
This is true since both terms are positive.

6. **Conclusion**:
Therefore, we conclude that:
\[ 2^{\sin \theta} + 2^{\cos \theta} \geq 2^{1 - \frac{1}{\sqrt{2}}} \]
holds for all real values of \( \theta \).