The inequality `2 ^("sin"^(2) theta )+2 ^(cos^(2) theta ) ge 2 sqrt( 2)` holds for all real `theta`

Answer

To solve the inequality \( 2^{\sin^2 \theta} + 2^{\cos^2 \theta} \geq 2\sqrt{2} \), we can follow these steps:

### Step 1: Use the identity for sine and cosine
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Let's denote \( x = \sin^2 \theta \). Then, \( \cos^2 \theta = 1 - x \).

### Step 2: Rewrite the inequality
Substituting \( x \) into the inequality gives us:
\[ 2^x + 2^{1-x} \geq 2\sqrt{2} \]

### Step 3: Simplify the expression
We can rewrite \( 2^{1-x} \) as \( \frac{2}{2^x} \):
\[ 2^x + \frac{2}{2^x} \geq 2\sqrt{2} \]

### Step 4: Let \( y = 2^x \)
Now, substituting \( y = 2^x \), we have:
\[ y + \frac{2}{y} \geq 2\sqrt{2} \]

### Step 5: Multiply through by \( y \)
To eliminate the fraction, multiply the entire inequality by \( y \) (noting that \( y > 0 \)):
\[ y^2 + 2 \geq 2\sqrt{2}y \]

### Step 6: Rearrange the inequality
Rearranging gives us:
\[ y^2 - 2\sqrt{2}y + 2 \geq 0 \]

### Step 7: Analyze the quadratic
The expression \( y^2 - 2\sqrt{2}y + 2 \) is a quadratic in \( y \). We can find its discriminant to determine if it has real roots:
\[ D = (2\sqrt{2})^2 - 4 \cdot 1 \cdot 2 = 8 - 8 = 0 \]
Since the discriminant is zero, the quadratic has one real root.

### Step 8: Find the root
Using the quadratic formula:
\[ y = \frac{2\sqrt{2} \pm 0}{2} = \sqrt{2} \]
Thus, the quadratic can be expressed as:
\[ (y - \sqrt{2})^2 \geq 0 \]
This is true for all \( y \) since squares are always non-negative.

### Step 9: Conclusion
Since \( y = 2^{\sin^2 \theta} \) can take any positive value, the inequality holds for all real \( \theta \).