International Mathematical Olympiad

We first prove that, for $x>0$, $$\begin{array}{c}{a}_i(x+1{)}^{1/a_i}⩽x+a_i,\end{array}\text{\hspace{1em}(1)}$$ 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 $$\frac{(x+1)+\stackrel{a_i-1\text{ ones }}{\overbrace{1+1+\cdots +1}}}{a_i}⩾\sqrt[a_i]{(x+1)\cdot 1^{a_i-1}}⟹a_i(x+1{)}^{1/a_i}⩽x+a_i.$$ Since $x+1>1$, the inequality is strict for $a_i>1$. Multiplying the inequalities (1) for $i=1,2,\dots ,n$ yields $$\prod _{i=1}^n{a}_i(x+1{)}^{1/a_i}⩽\prod _{i=1}^n\left(x+a_i\right)⟺M(x+1{)}^{{\sum }_{i=1}^{n}1/a_i}-\prod _{i=1}^n\left(x+a_i\right)⩽0⟺P(x)⩽0$$ with equality iff $a_i=1$ for all $i\in \{1,2,\dots ,n\}$. 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.

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Alternative solution.

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 $a_j⩾1$ for all $j$ and $a_j>1$ for some $j$ (since $a_1{a}_2\dots a_n=M>1$ ), 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 $k<r⩽n=\mathrm{deg}(P)$. For $0⩽r⩽k$, the coefficient of $x^r$ in $P(x)$ is $$M\cdot \left(\genfrac{}{}{0ex}{}{k}{r}\right)-\sum _{1⩽i_1<i_2<\cdots <i_{n-r}⩽n}{a}_{i_1}{a}_{i_2}\cdots a_{i_{n-r}}=a_1{a}_2\cdots a_n\cdot \left(\genfrac{}{}{0ex}{}{k}{r}\right)-\sum _{1⩽i_1<i_2<\cdots <i_{n-r}⩽n}{a}_{i_1}{a}_{i_2}\cdots a_{i_{n-r}},$$ which is non-positive iff $$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{k}{r}\right)⩽\sum _{1⩽j_1<j_2<\cdots <j_r⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}}.\end{array}\text{\hspace{1em}(2)}$$ 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 $$\left(\genfrac{}{}{0ex}{}{k}{r+1}\right)=\frac{k-r}{r+1}\cdot \left(\genfrac{}{}{0ex}{}{k}{r}\right)⩽\frac{k-r}{r+1}.\sum _{1⩽j_1<j_2<\cdots <j_r⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}},$$ and it suffices to prove that $$\frac{k-r}{r+1}.\sum _{1⩽j_1<j_2<\cdots <j_r⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}}⩽\sum _{1⩽j_1<\cdots <j_r<j_{r+1}⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}{a}_{j_{r+1}}},$$ which is equivalent to $$\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_n}-r\right)\sum _{1⩽j_1<j_2<\cdots <j_r⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}}⩽(r+1)\sum _{1⩽j_1<\cdots <j_r<j_{r+1}⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}{a}_{j_{r+1}}}.$$ 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 $$\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_n}\right)\sum _{1⩽j_1<j_2<\cdots <j_r⩽n}\frac{1}{a_{j_1}{a}_{j_2}\cdots a_{j_r}}.$$ Hence all terms in the right hand side cancel out.