75th NMO Selection Tests

a. Let us denote $$a_k=\frac{x_k}{1+x_1+\cdots +x_k}\text{and}{b}_k=1-a_k=\frac{1+x_1+\cdots +x_{k-1}}{1+x_1+\cdots +x_k},1\le k\le n.$$ By the inequality of arithmetic and geometric means, we have $$b_1+b_2+\cdots +b_n\ge n\sqrt[n]{b_1{b}_2\cdots b_n}=\frac{n}{\sqrt[n]{1+x_1+\cdots +x_n}}=\frac{n}{\sqrt[n]{2}}.$$ Therefore, $$n\cdot m\le a_1+a_2+\cdots +a_n=n-(b_1+b_2+\cdots +b_n)\le n\left(1-\frac{1}{\sqrt[n]{2}}\right),$$ which implies $m\le 1-\frac{1}{\sqrt[n]{2}}$.

We will show that there exists a sequence $(x_1,x_2,\dots ,x_n)$ of positive real numbers summing to 1 such that $a_1=a_2=\cdots =a_n=1-\frac{1}{\sqrt[n]{2}}\stackrel{\text{def}}{=}{m}_0$, which implies that the greatest possible value of $m$ is $m_0$. The existence of such a sequence is equivalent to solving the system $$x_1+\cdots +x_n=1,\frac{x_1}{1+x_1}=\frac{x_i}{1+x_1+\cdots +x_i},\text{for all }i=2,\dots ,n.$$ For $i=2$, we get $x_2=a_1(1+x_1)$. For $i=3$, replacing $x_2$, we find $x_3=x_1(1+x_1{)}^2$. By induction on $i$, it follows that $x_i=x_1(1+x_1{)}^{i-1}$, for all $i\le n$. Since their sum is 1, we get $$1=x_1+x_1(1+x_1)+\cdots +x_1(1+x_1{)}^{n-1}=x_1\frac{(1+x_1{)}^n-1}{(1+x_1)-1}=(1+x_1{)}^n-1.$$ Hence, $(1+x_1{)}^n=2$, so $x_1=\sqrt[n]{2}-1$. Substituting back, $x_k=\sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)$, $k=1,2,\dots ,n$.

b. We will prove that the smallest possible value of $M$ is $M_0=1-\frac{1}{\sqrt[n]{2}}$, which is attained if and only if $a_1=a_2=\cdots =a_n$. Using the same reasoning as before, we have $a_1=a_2=\cdots =a_n$ if and only if $x_k=\sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)$, $k=1,2,\dots ,n$. Let ${\alpha }_k=\sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)$ for $1\le k\le n$, and consider a positive sequence $(x_1,x_2,\dots ,x_n)$, different from $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)$, with sum 1. Then there exists $k\in \{1,\dots ,n\}$ such that ${\alpha }_k<x_k$. Let $k_0$ be the smallest such index for which ${\alpha }_{k_0}<x_{k_0}$; then $x_i\le {\alpha }_i$, for all $1\le i\le k_0-1$. We deduce $$\begin{array}{rl}{a}_{k_0}& =\frac{x_{k_0}}{1+x_1+\cdots +x_{k_0}}=\frac{1}{\frac{1+x_1+\cdots +x_{k_0-1}}{x_{k_0}}+1}>\frac{1}{\frac{1+{\alpha }_1+\cdots +{\alpha }_{k_0-1}}{{\alpha }_{k_0}}+1}\\ & =\frac{{\alpha }_{k_0}}{1+{\alpha }_1+\cdots +{\alpha }_{k_0}}=M_0,\end{array}$$ therefore, $M=\max a_k\ge a_{k_0}>M_0$. In conclusion, $M_0=1-\frac{1}{\sqrt[n]{2}}$ is the minimal possible value of $M$.