Mediterranean Mathematics Competition

Suppose first $n\ge 4$. Then we have

$$\begin{gathered} \frac{a_1{a}_2\dots a_{k-1}{a}_{k+1}\dots a_n}{a_k+n-2}\le \frac{{\left(\frac{a_1+a_2+\cdots +a_{k-1}+a_{k+1}+\cdots +a_n}{n-1}\right)}^{n-1}}{a_k+n-2}< \\ <\frac{{\left(\frac{a_1+a_2+\cdots +a_n}{n-1}\right)}^{n-1}}{n-2}=\frac{1}{(n-2)(n-1{)}^{n-1}} \end{gathered}$$ and by addition of all these inequalities we get $$\begin{gathered} \frac{a_2{a}_3\dots a_n}{a_1+n-2}+\frac{a_1{a}_3\dots a_n}{a_2+n-2}+\frac{a_1{a}_2{a}_4\dots a_n}{a_3+n-2}+\cdots +\frac{a_1{a}_2\dots a_{n-1}}{a_n+n-2}\le \frac{n}{(n-2)(n-1{)}^{n-1}}\le \\ \le \frac{n}{(n-2)(n-1{)}^3} \end{gathered}$$ and so we need to prove $$\frac{n}{(n-2)(n-1{)}^3}\le \frac{1}{(n-1{)}^2},$$ which simplifies to give $$n\le (n-2)(n-1)\iff 0\le n^2-4n+2\iff 0\le n(n-4)+2$$ which is true if $n\ge 4$.

In the case $n=3$ we put $a_1=a$, $a_2=b$, $a_3=c$ and we need to prove $$\frac{bc}{1+a}+\frac{ca}{1+b}+\frac{ab}{1+c}\le \frac{1}{4}.$$ As we have $$\frac{bc}{1+a}=bc-\frac{abc}{1+a},\frac{ac}{1+b}=ac-\frac{abc}{1+b},\frac{ab}{1+c}=ab-\frac{abc}{1+c}$$ the inequality to prove is $$bc+ca+ab\le \frac{1}{4}+abc\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right).$$ But the inequality of the harmonic-arithmetical means allow to writing $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge 3\frac{3}{(a+1)(b+1)+(c+1)}=\frac{9}{4},$$ and so we need to prove $$bc+ca+ab\le \frac{1}{4}+abc\cdot \frac{9}{4}.$$ As this is symmetrical in a,b,c we can suppose $c\le \frac{1}{3}$; then $$bc+(1-b-c)c+(1-b-c)b\le \frac{1}{4}+\frac{9}{4}(1-b-c)bc$$ And this is equivalent to $$\begin{array}{rl}0\le 4b^2-9b^{2}c-9b{c}^2-4b+4c^2-4c+1& \iff \\ 0\le (4-9c)b^2-(9c^2-13c+4)b+(4c^2-4c+1)& \end{array}$$ In we put $$p(x)=(4-9c)x^2-(9c^2-13c+4)x+(4c^2-4c+1),$$ as $c\le \frac{1}{3}$, $4-9c\ge 1$; and the discriminant of the polynomial is $$D=(9c^2-13c+4{)}^2-4(4-9c)(4c^2-4c+1)=81c{\left(c-\frac{1}{3}\right)}^2\left(c-\frac{4}{9}\right)$$ and then, from $0<c\le \frac{1}{3}$ and $c-\frac{4}{9}<0$ we obtain $D\le 0$, and the case $n=3$ is proved. The equality holds for $a=b=c=\frac{1}{3}$.