# Functional inequality

1. Oct 5, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
Let $f,g$ be two real valued functions, defined on the segment $[a,b]$ and continuous on $[a,b]$, such that $0 < g < f$. Show there exist $\lambda > 0$ such that $(1+\lambda) g \le f$

2. Relevant equations

3. The attempt at a solution

Set $h = f/g$. Since $g\neq 0$, $h$ is continuous on $[a,b]$.
Therefore, $h$ is bounded on $[a,b]$ and reaches its bounds. Call $m = h(x_0)$ its lower bound. By construction, $1 < m \le h$, so $\lambda = m-1 > 0$ and $(1+\lambda) g \le f$. Is it correct ?

2. Oct 5, 2015

### RUber

Looks good to me. Then $(1+\lambda)g(x_0) = f(x_0)$ and is less than or equal to at all other points in the domain.

3. Oct 5, 2015

### geoffrey159

Thank you !

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