IMO2024 Shortlisted Problems

Solution 1. For every positive integer $n$, let $M_n=\max \left(a_1,a_2,\dots ,a_n\right)$. We first prove that $$\frac{3^{a_1}+3^{a_2}+\cdots +3^{a_n}}{{\left(2^{a_1}+2^{a_2}+\cdots +2^{a_n}\right)}^2}⩽{\left(\frac{3}{4}\right)}^{M_n}$$ For $i=1,2,\dots ,n$, from ${\left(\frac{3}{2}\right)}^{a_i}⩽{\left(\frac{3}{2}\right)}^{M_n}$ we can obtain $3^{a_i}⩽{\left(\frac{3}{4}\right)}^{M_n}\cdot 2^{M_n}\cdot 2^{a_i}$. By summing up over all $i$, $$\sum _{i=1}^n{3}^{a_i}⩽{\left(\frac{3}{4}\right)}^{M_n}\cdot 2^{M_n}\cdot \sum _{i=1}^n{2}^{a_i}⩽{\left(\frac{3}{4}\right)}^{M_n}\cdot {\left(\sum _{i=1}^n{2}^{a_i}\right)}^2$$ which is equivalent to the previous inequality. Now let $\mu ={\log }_{4/3}\frac{1}{\epsilon }$, so that $\mu$ is the positive real number with ${\left(\frac{3}{4}\right)}^{\mu }=\epsilon$. If there is an index $n$ such that $a_n>\mu$, then $M_n⩾a_n>\mu$, and hence $$\frac{3^{a_1}+3^{a_2}+\cdots +3^{a_n}}{{\left(2^{a_1}+2^{a_2}+\cdots +2^{a_n}\right)}^2}⩽{\left(\frac{3}{4}\right)}^{M_n}<{\left(\frac{3}{4}\right)}^{\mu }=\epsilon .$$ Otherwise we have $0<a_i⩽\mu$ for all positive integers $i$, so $$\frac{3^{a_1}+3^{a_2}+\cdots +3^{a_n}}{{\left(2^{a_1}+2^{a_2}+\cdots +2^{a_n}\right)}^2}⩽\frac{n\cdot 3^{\mu }}{(n\cdot 1{)}^2}=\frac{3^{\mu }}{n}$$ If $n>\lfloor \frac{3^{\mu }}{\epsilon }\rfloor$, this is less than $\epsilon$.

Solution 2. We will combine two upper bounds. First, start with the trivial estimate $$\frac{3^{a_1}+\cdots +3^{a_n}}{{\left(2^{a_1}+\cdots +2^{a_n}\right)}^2}⩽\frac{3^{a_1}+\cdots +3^{a_n}}{4^{a_1}+\cdots +4^{a_n}}$$ By applying Jensen's inequality to the convex function $x^{{\log }_{3}4}$ we get $$\frac{4^{a_1}+\cdots +4^{a_n}}{n}=\frac{{\left(3^{a_1}\right)}^{{\log }_{3}4}+\cdots +{\left(3^{a_n}\right)}^{{\log }_{3}4}}{n}⩾{\left(\frac{3^{a_1}+\cdots +3^{a_n}}{n}\right)}^{{\log }_{3}4}$$ so $$\frac{3^{a_1}+\cdots +3^{a_n}}{{\left(2^{a_1}+\cdots +2^{a_n}\right)}^2}⩽\frac{3^{a_1}+\cdots +3^{a_n}}{4^{a_1}+\cdots +4^{a_n}}⩽{\left(\frac{n}{3^{a_1}+\cdots +3^{a_n}}\right)}^{{\log }_{3}4-1}$$ Hence, the original inequality holds true whenever $$3^{a_1}+\cdots +3^{a_n}>{\left(\frac{1}{\epsilon }\right)}^{\frac{1}{{\log }_{3}4-1}}\cdot n$$ Second, trivially $$\frac{3^{a_1}+\cdots +3^{a_n}}{{\left(2^{a_1}+\cdots +2^{a_n}\right)}^2}⩽\frac{3^{a_1}+\cdots +3^{a_n}}{n^2}$$ so the original inequality is satisfied if $$3^{a_1}+\cdots +3^{a_n}<\epsilon \cdot n^2$$ If $n>{\left(\frac{1}{\epsilon }\right)}^{1+\frac{1}{{\log }_{3}4-1}}$ then ${\left(\frac{1}{\epsilon }\right)}^{\frac{1}{{\log }_{3}4-1}}\cdot n<\epsilon \cdot n^2$, and therefore at least one of the two conditions is satisfied.

Solution 3. Define $C={\log }_{4/3}\frac{2}{\epsilon }$, so that if $a_i>C$ then $3^{a_i}<\frac{\epsilon }{2}\cdot 4^{a_i}$. We divide the sequence into "small" and "large" terms by how they compare to $C$ : let $${\mathcal{S}}_n=\left\{i⩽n\mid a_i⩽C\right\}\text{ and }{\mathcal{L}}_n=\left\{i⩽n\mid a_i>C\right\}$$ Then the original inequality is equivalent to $$\frac{{\sum }_{i\in {\mathcal{S}}_n}{3}^{a_i}}{{\left({\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}+{\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}\right)}^2}+\frac{{\sum }_{i\in {\mathcal{L}}_n}{3}^{a_i}}{{\left({\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}+{\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}\right)}^2}<\frac{\epsilon }{2}+\frac{\epsilon }{2}$$ If ${\mathcal{L}}_n$ is nonempty, we have $$\frac{{\sum }_{i\in {\mathcal{L}}_n}{3}^{a_i}}{{\left({\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}+{\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}\right)}^2}<\frac{\epsilon }{2}\cdot \frac{{\sum }_{i\in {\mathcal{L}}_n}{4}^{a_i}}{{\left({\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}\right)}^2}⩽\frac{\epsilon }{2}$$ and this also trivially holds when ${\mathcal{L}}_n$ is empty (in which case the LHS is zero). Now suppose that $n⩾\frac{2}{\epsilon }{\left(\frac{3}{2}\right)}^C$. Note that $3^{a_i}⩽{\left(\frac{3}{2}\right)}^C{2}^{a_i}$ for $i\in {\mathcal{S}}_n$, so we have $$\frac{{\sum }_{i\in {\mathcal{S}}_n}{3}^{a_i}}{{\left({\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}+{\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}\right)}^2}⩽\frac{{\left(\frac{3}{2}\right)}^C{\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}}{{\left({\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}+{\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}\right)}^2}⩽\frac{{\left(\frac{3}{2}\right)}^C}{{\sum }_{i\in {\mathcal{S}}_n}{2}^{a_i}+{\sum }_{i\in {\mathcal{L}}_n}{2}^{a_i}}<\frac{{\left(\frac{3}{2}\right)}^C}{n}⩽\frac{\epsilon }{2},$$ so we have the original inequality.

Solution 4. For every index $i=1,2,\dots ,n$, apply the weighted AM-GM inequality to numbers $2^{a_i}$ and $(n-1)$ with weights ${\log }_2\frac{3}{2}\approx 0.585$ and ${\log }_2\frac{4}{3}\approx 0.415$ as $$\begin{array}{c}{2}^{a_1}+2^{a_2}+\cdots +2^{a_n}⩾2^{a_i}+(n-1)\\ >{\log }_2\frac{3}{2}\cdot 2^{a_i}+{\log }_2\frac{4}{3}\cdot (n-1)⩾{\left(2^{a_i}\right)}^{{\log }_2\frac{3}{2}}\cdot (n-1{)}^{{\log }_2\frac{4}{3}}\\ ={\left(\frac{3}{2}\right)}^{a_i}\cdot (n-1{)}^{{\log }_2\frac{4}{3}}>{\left(\frac{3}{2}\right)}^{a_i}\cdot (n-1{)}^{2/5}.\end{array}$$ By summing up for $i=1,2,\dots ,n$, $${\left(2^{a_1}+\cdots +2^{a_n}\right)}^2=\sum _{i=1}^n{2}^{a_i}\left(2^{a_1}+2^{a_2}+\cdots +2^{a_n}\right)>(n-1{)}^{2/5}\sum _{i=1}^n{3}^{a_i}$$ so $$\frac{3^{a_1}+3^{a_2}+\cdots +3^{a_n}}{{\left(2^{a_1}+2^{a_2}+\cdots +2^{a_n}\right)}^2}<\frac{1}{(n-1{)}^{2/5}}$$ If $n⩾{\left(\frac{1}{\epsilon }\right)}^{5/2}+1$ then $\frac{1}{(n-1{)}^{2/5}}<\epsilon$.