International Mathematical Olympiad

We first prove that, for $x>0$,
$$\begin{equation*} a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i}, \tag{1} \end{equation*}$$
with equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$.
If $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields
$$\frac{(x+1)+\overbrace{1+1+\cdots+1}^{a_{i}-1 \text{ ones }}}{a_{i}} \geqslant \sqrt[a_{i}]{(x+1) \cdot 1^{a_{i}-1}} \Longrightarrow a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i} .$$
Since $x+1>1$, the inequality is strict for $a_{i}>1$.
Multiplying the inequalities (1) for $i=1,2, \ldots, n$ yields
$$\prod_{i=1}^{n} a_{i}(x+1)^{1 / a_{i}} \leqslant \prod_{i=1}^{n}\left(x+a_{i}\right) \Longleftrightarrow M(x+1)^{\sum_{i=1}^{n} 1 / a_{i}}-\prod_{i=1}^{n}\left(x+a_{i}\right) \leqslant 0 \Longleftrightarrow P(x) \leqslant 0$$
with equality iff $a_{i}=1$ for all $i \in\{1,2, \ldots, n\}$. But this implies $M=1$, which is not possible. Hence $P(x)<0$ for all $x \in \mathbb{R}^{+}$, and $P$ has no positive roots.

We will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$.
Indeed, since $a_{j} \geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \ldots a_{n}=M>1$ ), we have $k=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<n$, so the coefficient of $x^{n}$ in $P(x)$ is $-1<0$. Moreover, the coefficient of $x^{r}$ in $P(x)$ is negative for $k<r \leqslant n=\operatorname{deg}(P)$.
For $0 \leqslant r \leqslant k$, the coefficient of $x^{r}$ in $P(x)$ is
$$M \cdot\binom{k}{r}-\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{n-r} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{n-r}}=a_{1} a_{2} \cdots a_{n} \cdot\binom{k}{r}-\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{n-r} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{n-r}},$$
which is non-positive iff
$$\begin{equation*} \binom{k}{r} \leqslant \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} . \tag{2} \end{equation*}$$
We will prove (2) by induction on $r$. For $r=0$ it is an equality because the constant term of $P(x)$ is $P(0)=0$, and if $r=1$, (2) becomes $k=\sum_{i=1}^{n} \frac{1}{a_{i}}$. For $r>1$, if (2) is true for a given $r<k$, we have
$$\binom{k}{r+1}=\frac{k-r}{r+1} \cdot\binom{k}{r} \leqslant \frac{k-r}{r+1} . \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}},$$
and it suffices to prove that
$$\frac{k-r}{r+1} . \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \leqslant \sum_{1 \leqslant j_{1}<\cdots<j_{r}<j_{r+1} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}},$$
which is equivalent to
$$\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-r\right) \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \leqslant(r+1) \sum_{1 \leqslant j_{1}<\cdots<j_{r}<j_{r+1} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}} .$$
Since there are $r+1$ ways to choose a fraction $\frac{1}{a_{j_{i}}}$ from $\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}}$ to factor out, every term $\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}}$ in the right hand side appears exactly $r+1$ times in the product
$$\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} .$$
Hence all terms in the right hand side cancel out.