Baltic Way Shortlist, 2023

For $a, b \in \mathbb{R}_{>0}$ we consider the function $f: \mathbb{R}_{\ge 0} \to \mathbb{R}$ with $f(x) = b x^3 - a x$ for all $x \in \mathbb{R}_{\ge 0}$.
For $x \ge 0$ we have $b(x - \sqrt{\frac{a}{3b}})^2 (x + 2\sqrt{\frac{a}{3b}}) \ge 0$ which is equivalent to
$$f(x) = b x^3 - a x \geq -\frac{2a}{3}\sqrt{\frac{a}{3b}} = f\left(\sqrt{\frac{a}{3b}}\right) = \min(f).$$
For $i \in \{1, \dots, n\}$ consider the functions $f_i: \mathbb{R}_{\ge 0} \to \mathbb{R}$ with $f_i(x) = b_i x^3 - a_i x$ for all $x \in \mathbb{R}_{\ge 0}$, and put $f = f_1 + f_2 + \dots + f_n$. Note that
$$f(x) = (b_1 + b_2 + \dots + b_n)x^3 - (a_1 + a_2 + \dots + a_n)x$$
for all $x \in \mathbb{R}_{\ge 0}$. It follows that
$$\min(f_i) = -\frac{2}{3\sqrt{3}} \cdot a_i \sqrt{\frac{a_i}{b_i}}$$
for all $i \in \{1, \dots, n\}$ and
$$\min(f) = -\frac{2}{3\sqrt{3}} \cdot (a_1 + a_2 + \dots + a_n) \sqrt{\frac{a_1 + a_2 + \dots + a_n}{b_1 + b_2 + \dots + b_n}}.$$
The desired inequality follows from the inequality
$$\min(f_1) + \min(f_2) + \dots + \min(f_n) \le \min(f_1 + f_2 + \dots + f_n)$$