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

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 \( a, b > 0 \).