Mediterranean Mathematics Competition

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

$$\frac{a_1 a_2 \dots a_{k-1} a_{k+1} \dots a_n}{a_k + n - 2} \le \frac{\left( \frac{a_1 + a_2 + \dots + a_{k-1} + a_{k+1} + \dots + a_n}{n-1} \right)^{n-1}}{a_k + n - 2} < \\ < \frac{\left( \frac{a_1 + a_2 + \dots + a_n}{n-1} \right)^{n-1}}{n-2} = \frac{1}{(n-2)(n-1)^{n-1}}$$
and by addition of all these inequalities we get
$$\frac{a_2 a_3 \dots a_n}{a_1 + n-2} + \frac{a_1 a_3 \dots a_n}{a_2 + n-2} + \frac{a_1 a_2 a_4 \dots a_n}{a_3 + n-2} + \dots + \frac{a_1 a_2 \dots a_{n-1}}{a_n + n-2} \le \frac{n}{(n-2)(n-1)^{n-1}} \le \\ \le \frac{n}{(n-2)(n-1)^3}$$
and so we need to prove
$$\frac{n}{(n-2)(n-1)^3} \le \frac{1}{(n-1)^2},$$
which simplifies to give
$$n \le (n-2)(n-1) \quad \Leftrightarrow \quad 0 \le n^2 - 4n + 2 \quad \Leftrightarrow \quad 0 \le n(n-4) + 2$$
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
$$\frac{bc}{1+a} = bc - \frac{abc}{1+a}, \quad \frac{ac}{1+b} = ac - \frac{abc}{1+b} , \quad \frac{ab}{1+c} = ab - \frac{abc}{1+c}$$
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
$$\begin{aligned} 0 \le 4b^2 - 9b^2c - 9bc^2 - 4b + 4c^2 - 4c + 1 &\Leftrightarrow \\ 0 \le (4 - 9c)b^2 - (9c^2 - 13c + 4)b + (4c^2 - 4c + 1) \end{aligned}$$
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 $a = b = c = \frac{1}{3}$.