Baltic Way 2023 Shortlist

For $a,b\in {\mathbb{R}}_{>0}$ we consider the function $f:{\mathbb{R}}_{\ge 0}\to \mathbb{R}$ with $f(x)=b{x}^3-ax$ 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 $$f(x)=b{x}^3-ax\ge -\frac{2a}{3}\sqrt{\frac{a}{3b}}=f\left(\sqrt{\frac{a}{3b}}\right)=\min (f).$$ 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 $f=f_1+f_2+\cdots +f_n$. Note that $$f(x)=(b_1+b_2+\cdots +b_n)x^3-(a_1+a_2+\cdots +a_n)x$$ 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 $$\min (f)=-\frac{2}{3\sqrt{3}}\cdot (a_1+a_2+\cdots +a_n)\sqrt{\frac{a_1+a_2+\cdots +a_n}{b_1+b_2+\cdots +b_n}}.$$ The desired inequality follows from the inequality $$\min (f_1)+\min (f_2)+\cdots +\min (f_n)\le \min (f_1+f_2+\cdots +f_n)$$