A polynomial problem of an indian olympiad Let $a_1,a_2,.....a_n$, $k$ and $M$ be positive integers such that:
1. $\frac{1}{a_1}+\frac{1}{a_2}+....\frac{1}{a_n}=k$
2. $a_1a_2.....a_n=M$
If $M>1$, prove that the polynomial $P(x)=M(x+1)^k−(x+a_1)(x+a_2).....(x+a_n)$ has no positives roots.
I have no hint how to solve it. Please try it.
 A: At first we have:
$$P(x) = 0 \Longrightarrow M (x+1)^k = (x+a_1)...(x+a_n) \Longrightarrow \prod_{i=1}^{n}a_i(x+1)^\frac{1}{a_i} = \prod_{i=1}^{n}(x+a_i)$$
Now we show for every $i$:
$$1 \leqslant a_i(x+1)^\frac{1}{a_i} \leqslant (x+a_i)$$
First way:
If take derive from both sides we have:
$$(x+1)^{\frac{1}{a_i}-1} \leqslant 1$$
Since $a_i \geqslant 1 \ \& \ x \geqslant 0$ the last inequality hold. And the equality hold iff $x=0$ or $a_i=1$. Also this two function meet the other at $x=0$:
$$a_i(x+1)^\frac{1}{a_i} = a_i = (x+a_i)$$
So the first inequality also hold for every positive $x$.
Second way: power sides:
$$a_i^{a_i}(x+1) \leqslant (x+a_i)^{a_i}$$
$$a_i^{a_i}x+a_i^{a_i} \leqslant a_i^{a_i} + a_i^{a_i}x + \sum_{j=2}^{a_i}b_jx^j$$
Since all parameter are non negative the inequality hold. And the equality hold iff $x=0$ or $a_i=1$. so the first equality also hold for every positive $x$.
After proof of inequality in every of two way we have:
By $M>1$ we have at least one $l$ that $a_l>1$, so we have:
$$a_l(x+1)^\frac{1}{a_l} < (x+a_l)$$
Since There is no $zero$ factor we have:
$$\prod_{i=1}^{n}a_i(x+1)^\frac{1}{a_i} < \prod_{i=1}^{n}(x+a_i)$$
So no $positive$ roots allowed.
