Problem 1362

1. Understanding the Problem:
We are given a convex function $f : (0, +\infty) \to \mathbb{R}$ and positive constants $\alpha, \beta, \gamma$. We need to prove the inequality involving integrals of $f$ over different intervals.

2. Convex Function Property:
Recall that a function $f$ is convex if for any $x_1, x_2 \in (0, +\infty)$ and $\lambda \in [0, 1]$:
$$f(\lambda x_1 + (1-\lambda) x_2) \leq \lambda f(x_1) + (1-\lambda) f(x_2).$$
This property will be useful in our proof.

3. Averaging Property of Integrals:
For a convex function $f$, the average value of $f$ over an interval is less than or equal to the value of $f$ at the midpoint of the interval. Formally, for any interval $[a, b]$:
$$\frac{1}{b-a} \int_a^b f(x) \, dx \leq f\left(\frac{a+b}{2}\right).$$

4. Applying the Convexity:
We need to show:
$$\frac{1}{6\alpha}\int_0^{6\alpha} f(x) \, dx + \frac{1}{6\beta}\int_0^{6\beta} f(x) \, dx + \frac{1}{6\gamma}\int_0^{6\gamma} f(x) \, dx \geq \frac{1}{3\alpha + 2\beta + \gamma}\int_0^{3\alpha + 2\beta + \gamma} f(x) \, dx + \frac{1}{\alpha + 3\beta + 2\gamma}\int_0^{\alpha + 3\beta + 2\gamma} f(x) \, dx + \frac{1}{2\alpha + \beta + 3\gamma}\int_0^{2\alpha + \beta + 3\gamma} f(x) \, dx.$$

5. Using Jensen's Inequality:
Jensen's inequality for integrals states that for a convex function $f$ and a probability measure $\mu$:
$$f\left(\int x \, d\mu(x)\right) \leq \int f(x) \, d\mu(x).$$
Applying this to our problem, we consider the weighted averages of the integrals.

6. Combining the Integrals:
We need to show that the weighted average of the integrals on the left-hand side is greater than or equal to the weighted average of the integrals on the right-hand side. This follows from the convexity of $f$ and the fact that the intervals on the left-hand side are larger, thus the average value of $f$ over these intervals is greater.

7. Conclusion:
By the properties of convex functions and Jensen's inequality, the given inequality holds.

$\blacksquare$