If $f(x), g(x)$ and $h(x)$ are continuous and positive functions such that $f(x) + g(x) + h(x) = \sqrt{f(x)g(x)} + \sqrt{g(x)h(x)} + \sqrt{h(x)f(x)}$, then $\int (f(x) + g(x) - 2h(x))dx$ is/are
Step-by-Step Solution
Key Concept: The constraint equation f(x) + g(x) + h(x) = √(f(x)g(x)) + √(g(x)h(x)) + √(h(x)f(x)) is satisfied only when f(x) = g(x) = h(x), which can be proven using AM-GM inequality: each pair's inequality becomes an equality only when the terms are equal.
Given $f(x) + g(x) + h(x)$ and the condition that $\sqrt{f(x) \cdot g(x)} + \sqrt{g(x) \cdot h(x)} + \sqrt{h(x) \cdot f(x)} = \sqrt{f(x) + g(x) + h(x)}$, we expand and analyze when this holds. This equation is satisfied only when $f(x) = g(x) = h(x)$, which can be verified by noting that the sum of squares $\frac{1}{2}[(\sqrt{f(x)} - \sqrt{g(x)})^2 + (\sqrt{g(x)} - \sqrt{h(x)})^2 + (\sqrt{h(x)} - \sqrt{f(x)})^2] = 0$ implies all three terms are equal. Therefore, $f(x) + g(x) - 2h(x) = 0$.
Correct Answer: 1,2,3