2017 IMO SL #A1

Let $a_1,a_2,\ldots a_n,k$, and $M$ be positive integers such that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=k\quad\text{and}\quad a_1a_2\cdots a_n=M.$$If $M>1$, prove that the polynomial $$P(x)=M(x+1)^k-(x+a_1)(x+a_2)\cdots (x+a_n)$$has no positive roots.



Assume for the sake of contradiction that $P(x)$ does have positive root(s).

Using the weighted AM-GM inequality, note that $$x + a_i = (x+1) + (a_i-1 \cdot 1) \ge a_i (x+1)^{\frac{1}{a_i}}.$$ 

Multiplying $x+a_i \ge  a_i (x+1)^{\frac{1}{a_i}}$ over all $i = 1 \to n$, we get $$\prod_{i=1}^{n} (x+a_i) \ge a_i\dots a_n(x+1)^{\frac{1}{a_1}+\dots+\frac{1}{a_n}}= M(x+1)^k. \space \space \space \space \space \space \space \space (*)$$ Since $$P(x)=M(x+1)^k-(x+a_1)(x+a_2)\cdots (x+a_n)$$, because of $(*)$ we know that $P(x) \leq 0$. However, recall that we assumed $P(x)$ has positive roots. So we must have equality. But we can only have equality when $x+1=1 \implies x=0$, which is a contradiction to the fact that we need the roots to be positive. 

We are done. $\square$



Comments

Popular posts from this blog

1995 IMO #2

Inequality problem I made