Croatian Mathematical Olympiad

Claim. For any positive integer $a$ and real number $x>0$ the following inequality holds: $$a(x+1{)}^{\frac{1}{a}}\le x+a,$$ where the equality is satisfied if and only if $a=1$.

Proof. If $a=1$, we easily see that the equality holds for all real numbers $x>0$. Let us assume that $a>1$, and let us apply the inequality between arithmetic and geometric means for $x+1$ and $a-1$ copies of $1$: $$\frac{(x+1)+\stackrel{a-1}{\overbrace{1+1+\cdots +1}}}{a}\ge \sqrt[a]{(x+1)\cdot 1^{a-1}},$$ or equivalently, $$a(x+1{)}^{\frac{1}{a}}\le x+a.$$ Notice that the equality is attained only if the equation $x+1=1$ is satisfied, which is impossible due to the assumption $x>0$. $\square$

Let us now prove the original problem, using the above claim. We have $$\begin{array}{rl}M(x+1{)}^k& =a_1(x+1{)}^{\frac{1}{a_1}}{a}_2(x+1{)}^{\frac{1}{a_2}}\cdots a_n(x+1{)}^{\frac{1}{a_n}}\\ & \le (x+a_1)(x+a_2)\cdots (x+a_n),\end{array}$$ where equality holds in case $a_1=a_2=\cdots =a_n=1$. However, $a_1{a}_2\cdots a_n=M>1$ so we conclude that this is not the case. Therefore, we conclude that the original equation has no solutions in positive real numbers.