Undergraduate contests Jozsef Wildt International Math Competition 2019 Jozsef Wildt International Math Competition W. 48 Prove this integral inequality of the convex function Problem Statement Let f : ( 0 , + ∞ ) → R f : (0,+\infty) \to \mathbb{R} f : ( 0 , + ∞ ) → R a convex function and α , β , γ > 0 \alpha, \beta, \gamma > 0 α , β , γ > 0 . Then 1 6 α ∫ 0 6 α f ( x ) d x + 1 6 β ∫ 0 6 β f ( x ) d x + 1 6 γ ∫ 0 6 γ f ( x ) d x \frac{1}{6\alpha}\int \limits_0^{6\alpha}f(x)dx\ +\ \frac{1}{6\beta}\int \limits_0^{6\beta}f(x)dx\ +\ \frac{1}{6\gamma}\int \limits_0^{6\gamma}f(x)dx 6 α 1 0 ∫ 6 α f ( x ) d x + 6 β 1 0 ∫ 6 β f ( x ) d x + 6 γ 1 0 ∫ 6 γ f ( x ) d x ≥ 1 3 α + 2 β + γ ∫ 0 3 α + 2 β + γ f ( x ) d x + 1 α + 3 β + 2 γ ∫ 0 α + 3 β + 2 γ f ( x ) d x \geq \frac{1}{3\alpha +2\beta +\gamma}\int \limits_0^{3\alpha +2\beta +\gamma}f(x)dx\ +\ \frac{1}{\alpha +3\beta +2\gamma}\int \limits_0^{\alpha +3\beta +2\gamma}f(x)dx\ ≥ 3 α + 2 β + γ 1 0 ∫ 3 α + 2 β + γ f ( x ) d x + α + 3 β + 2 γ 1 0 ∫ α + 3 β + 2 γ f ( x ) d x + 1 2 α + β + 3 γ ∫ 0 2 α + β + 3 γ f ( x ) d x +\ \frac{1}{2\alpha +\beta +3\gamma}\int \limits_0^{2\alpha +\beta +3\gamma}f(x)dx + 2 α + β + 3 γ 1 0 ∫ 2 α + β + 3 γ f ( x ) d x