# Dorin Marghidanu's Tribute to Descartes

### Solution 1

Due to Descartes' Rule of Signs, all roots of the polynomial are nonpositive. Assume these are $-x_k,\,$ $k=1,\ldots,n,\,$ with all $x_k\ge 0.\,$ Since, by Viète's formulas, $\displaystyle \prod_{k=1}^nx_k=1,\,$ the all $x_k\gt 0.\,$ By the AM-GM inequality,

$\displaystyle P(1)=\prod_{k=1}^n(1+x_k)\ge\prod_{k=1}^n(2\sqrt{x_k})=2^n\sqrt{\prod_{k=1}^nx_k}=2^n.$

But $P(1)=2+\displaystyle \sum_{k=1}^{n-1}a_k,\,$ thus proving the result.

### Solution 2

Due to Descartes' Rule of Signs, all roots of the polynomial are negative. Let these be $-x_k,\,$ $k=1,\ldots,n,\,$ with all $x_k\gt 0.\,$ Thus,

(1)

$P(x)=(x+x_1)(x+x_2)\cdot\ldots\cdot(x+x_n).$

Using Viète's relations

(2)

$\displaystyle x_1\cdot x_2\cdot\ldots\cdot x_{n}=1.$

With the AM-GM inequality,

(3)

$\displaystyle k+x_i=\underbrace{1+\ldots+1}_{\text{k times}}+x_i\ge (k+1)\sqrt[k+1]{x_i}.$

Putting together (1) and (3),

(4)

$\displaystyle P(k)=\prod_{i=1}^n(k+x_i)\ge (k+1)^n\sqrt[k+1]{x_1\cdot\ldots\cdot x_n}=(k+1)^n.$

Letting in (4) $k=1\,$ delivers the result.

### Solution 3

We'll invoke a corollary of the Binomial Theorem,

$\displaystyle {n\choose 0}+{n\choose 1}+\ldots+{n\choose n}=2^n.$

Let $-x_i,\,$ $x_i\gt 0,\,$ $i=1,\ldots,n,\,$ be the roots of the polynomial. $x_1\cdot\ldots\cdot x_n=1.\,$ Further, Using Viète's relations, and the AM-GM inequality,

\displaystyle \begin{align} &a_{n-1}=\sum_{i=1}^nx_i\ge{n\choose 1}\sqrt[{n\choose 1}]{\prod_{i=1}^nx_i}={n\choose 1}\\ &a_{n-2}=\sum_{i=1,j=1}^nx_ix_j\ge{n\choose 2}\sqrt[{n\choose 2}]{\prod_{i=1,j=1}^nx_ix_j}={n\choose 2}\\ &\qquad\qquad\cdots\cdots\\ &a_{1}=\sum x_{i_1}x_{i_2}\cdot\ldots\cdot x_{i_{n-1}}\ge{n\choose n-1}\sqrt[{n\choose n-1}]{\prod_{i=1}^n x_{i_1}x_{i_2}\cdot\ldots\cdot x_{i_{n-1}}}={n\choose n-1}. \end{align}

Summing up and using the corollary of the Binomial Theorem we mentioned at the outset gives the desired inequality.

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution 1 is by Leo Giugiuc; Solution 2 is Dorin Marghidanu; Solution 3 is by Vaggelis Stamatiadis.