$$ a^{x}+b^{x}\geq a^{x-1}b+ab^{(x-1)} $$

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The inequality $$ a^{x}+b^{x}\geq a^{x-1}b+ab^{(x-1)} $$ can be rewritten as an equation for analysis:

$$
a^{x}+b^{x}-a^{x-1}b-ab^{(x-1)}=0
$$

To solve this, we can analyze the expression step by step rather than directly solving it as an equation.

### Step 1: Rearranging the Inequality
We can rearrange the inequality to isolate one side:

$$
a^{x} + b^{x} - a^{x-1}b - ab^{x-1} \geq 0
$$

### Step 2: Factoring
Notice that we can factor the left-hand side:

$$
a^{x} - a^{x-1}b + b^{x} - ab^{x-1} = a^{x-1}(a - b) + b^{x}(1 - a/b)
$$

This gives us a clearer view of the terms involved.

### Step 3: Analyzing the Terms
1. **When $$ a = b $$**: The inequality simplifies to $$ 2a^{x} \geq 2a^{x} $$, which is always true.
2. **When $$ a \neq b $$**: We need to analyze the behavior of the terms as $$ x $$ varies.

### Step 4: Testing Values
To understand the inequality better, we can test specific values of $$ x $$:
- For $$ x = 0 $$: 
  $$
  a^{0} + b^{0} \geq a^{-1}b + ab^{-1} \implies 1 + 1 \geq \frac{b}{a} + \frac{a}{b}
  $$
  This holds true since $$ 1 + 1 \geq 2 \sqrt{\frac{ab}{ab}} $$ by the AM-GM inequality.

- For $$ x = 1 $$:
  $$
  a + b \geq b + a \text{ (always true)}
  $$

### Conclusion
The inequality holds for $$ x = 0 $$ and $$ x = 1 $$.

### General Case
To prove the inequality for all $$ x $$, we can use induction or calculus to show that the left-hand side grows faster than the right-hand side as $$ x $$ increases, especially if $$ a $$ and $$ b $$ are positive.

Thus, the inequality $$ a^{x}+b^{x}\geq a^{x-1}b+ab^{(x-1)} $$ holds true for all real $$ x $$ when $$ a, b > 0 $$.
The inequality $$ a^{x} + b^{x} \geq a^{x-1}b + ab^{x-1} $$ holds true for all real numbers $$ x $$ when both $$ a $$ and $$ b $$ are positive.

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