Key Concepts and Formulas

• AM-GM Inequality: For any two non-negative real numbers $A$ and $B$, the arithmetic mean is greater than or equal to the geometric mean: $\frac{A+B}{2} \ge \sqrt{AB}$, which implies $A+B \ge 2\sqrt{AB}$. Equality holds if and only if $A=B$.
• Properties of Exponents: For any positive base $a$ and real numbers $m, n$: $a^m \cdot a^n = a^{m+n}$.
• Logarithms: For $a>0, a \ne 1$, the equation $a^x = k$ has a unique real solution $x = \log_a k$ for any $k>0$.

Step-by-Step Solution

1. Identify the Function and Domain: The given function is $f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$, where $a, x \in \mathbb{R}$ and $a > 0$. We need to find the minimum value of this function.

2. Recognize Applicability of AM-GM Inequality: The function $f(x)$ is a sum of two terms. Let $A = {a^{{a^x}}}$ and $B = {a^{1 - {a^x}}}$. Since $a > 0$, any real power of $a$ is positive. Therefore, $a^x > 0$ for all $x \in \mathbb{R}$. Consequently, $A = a^{a^x} > 0$ and $B = a^{1-a^x} > 0$. Since both terms are positive, the AM-GM inequality can be applied.

3. Apply the AM-GM Inequality: Using the AM-GM inequality for $A$ and $B$: $f(x) = A + B \ge 2\sqrt{AB}$ Substitute the expressions for $A$ and $B$: ${a^{{a^x}}} + {a^{1 - {a^x}}} \ge 2\sqrt{{a^{{a^x}}} \cdot {a^{1 - {a^x}}}}$

4. Simplify the Product Term: Simplify the expression inside the square root using the exponent rule $a^m \cdot a^n = a^{m+n}$: ${a^{{a^x}}} \cdot {a^{1 - {a^x}}} = {a^{{a^x} + (1 - {a^x})}}$ The exponents simplify: ${a^x} + (1 - {a^x}) = a^x + 1 - a^x = 1$ So, the product becomes: ${a^{{a^x}}} \cdot {a^{1 - {a^x}}} = a^1 = a$

5. Determine the Lower Bound of f(x): Substitute the simplified product back into the AM-GM inequality: $f(x) \ge 2\sqrt{a}$ This shows that the minimum value of $f(x)$ is at least $2\sqrt{a}$.

6. Check for Equality Condition: For $f(x)$ to achieve its minimum value of $2\sqrt{a}$, the equality condition in the AM-GM inequality must hold. Equality occurs when $A = B$: ${a^{{a^x}}} = {a^{1 - {a^x}}}$ Since the bases are equal and positive (and assuming $a \ne 1$), their exponents must be equal: ${a^x} = 1 - {a^x}$ Solving for $a^x$: $2{a^x} = 1$ ${a^x} = \frac{1}{2}$ We need to confirm if there exists a real value of $x$ for which $a^x = \frac{1}{2}$, given $a > 0$.

   • If $a=1$, then $1^x = 1/2$ has no solution. However, if $a=1$, $f(x) = 1^{1^x} + 1^{1-1^x} = 1^1 + 1^{1-1} = 1+1^0 = 1+1=2$. Our derived minimum $2\sqrt{a}$ for $a=1$ is $2\sqrt{1}=2$. So the equality condition holds even for $a=1$ if we interpret $a^x=1/2$ as the condition derived from $a^{a^x} = a^{1-a^x}$. If $a=1$, then $1^{1^x} = 1^{1-1^x} \implies 1=1$, which is always true. So the equality holds for all $x$.
   • If $a \ne 1$ and $a > 0$, the equation $a^x = \frac{1}{2}$ has a unique real solution $x = \log_a \left(\frac{1}{2}\right) = -\log_a 2$. Since $a > 0$ and $a \ne 1$, $\log_a 2$ is a well-defined real number, so such an $x$ always exists.

7. Conclusion: Since the equality condition $a^x = \frac{1}{2}$ is achievable for some real value of $x$ (for any $a>0$), the minimum value of $f(x)$ is exactly $2\sqrt{a}$.

Common Mistakes & Tips

• Forgetting the Equality Condition: Simply applying AM-GM and getting $2\sqrt{a}$ is not enough. You must verify that the equality condition ($A=B$) can actually be met for some value of $x$.
• Handling the Base $a=1$: While $a^x = 1/2$ has no solution if $a=1$, the original equality $a^{a^x} = a^{1-a^x}$ becomes $1^{1^x} = 1^{1-1^x}$, which simplifies to $1=1$ and is true for all $x$. Thus, the equality condition holds for $a=1$ as well, and $f(x)=2$ for $a=1$, which matches $2\sqrt{a}=2\sqrt{1}=2$.
• Domain of $a^x$: Remember that for $a>0$, $a^x$ can take any positive real value. This is crucial for ensuring that $a^x = 1/2$ is solvable.

Summary

The function $f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$ is a sum of two positive terms. By applying the AM-GM inequality, we establish a lower bound of $2\sqrt{a}$. We then confirm that this lower bound is achievable by finding the condition for equality ($a^x = 1/2$) and verifying that a real value of $x$ exists to satisfy this condition for any $a>0$. Thus, the minimum value of the function is $2\sqrt{a}$.

The final answer is $\boxed{2\sqrt a}$.