- Let $ r_1,r_2,....,r_n$ be roots of polynomial $ f(x)$ and therefore real numbers. We have $ f(x) = \prod_{i = 1}^{n}(x - r_i)$ - By Viet's formulas we have $ n = - \sum_{i = 1}^{n} r_i$ and $ \frac {n(n - 1)}{2} = \frac {1}{2}\cdot\sum_{i = 1}^{n} r_i((\sum_{j = 1}^{n}r_j) - r_i)$ - Denote $ \frac {1}{2}\cdot\sum_{i = 1}^{n} r_i((\sum_{j = 1}^{n}r_j) - r_i)$ as $ A$ - if $ r_i$ are real numbers than we have by rearanging inequalitiy $ \sum_{i = 1}^{n}r_{i}^2\geq r_1r_2 + r_2r_3 + ... + r_nr_1$ $ \sum_{i = 1}^{n}r_{i}^2\geq r_1r_3 + r_2r_4 + ... + r_nr_2$ . . . $ \sum_{i = 1}^{n}r_{i}^2\geq r_1r_n + r_2r_1 + ... + r_nr_{n - 1}$ - adding all this together and deviding both sides by $ n - 1$ we get $ \sum_{i = 1}^{n}r_{i}^2\geq \frac {2A}{n - 1}$ - now we have $ n^2 = ( - n)^2 = (\sum_{i = 1}^{n}r_i)^2 = 2A + \sum_{i = 1}^{n}r_{i}^2\geq 2A + \frac {2A}{n - 1} = 2A\frac {n}{n - 1}$ which is obviously equivalent to $ \frac {n(n - 1)}{2}\geq A$ with equality at $ r_1 = r_2 = ... = r_n$ - because $ r_i$ are real numbers and because of Viet's formulas we know that $ A = \frac {n(n - 1)}{2}$, yielding $ r_1 = r_2 = ...r_n = - 1$ -giving $ f(x) = (x + 1)^n$ and $ g(x) = \sum_{i = 3}^{n}\binom{n}{i}x^{n - i}$ as the only solution for $ g(x)$ when $ n\geq3$ and $ g(x)=0$ if $ n < 3$

Rewire $ f(x) = x^n + nx^{n - 1} + \frac {n(n - 1)}2 x^{n - 2} + g(x) = (x - x_1)(x - x_2)(x - x_3)...(x - x_n)$. All roots of this equation real numbers. * By theory Viete's , we have : $ \sum_{i = 1}^{n} x_i = - n$ and $ \sum_{i \not = j}^{n} x_ix_j = \frac {n(n - 1)}{2}$ * But by other hand , we also have : $ \sum_{i = 1}^{n} x_i^2 = \sum_{i = 1}^{n} - \sum_{i \not = j}^{n} x_ix_j = n$ And by inequality Cauchy- Schwarz, we have : $ n^2 = n( \sum_{i = 1}^{n} x_i^2) \geq (\sum_{i = 1}^{n}x_i)^2 = n^2$ Occur equal only and if only $ x_i = - 1.$ hence : $ g(x) = f(x) - x^n - nx^{n - 1} - \frac {n(n - 1)}2 x^{n - 2}$ Conclude that $ g(x) = (x + 1)^n - x^n - nx^{n - 1} - \frac {n(n - 1)}2 x^{n - 2}$

Since $ f(x)$ is monic, let \[ f(x) = \prod_{i=1}^n (x+r_i) \] be the factorization of $ f$. We are told that the roots of $ f$, and hence the $ r_i$, are all real. Also, we know from Viète's relations that \[ \sum_i r_i = n, \qquad \sum_{i\neq j} r_ir_j = \frac{n(n-1)}{2} . \] Hence \[ \sum_i r_i^2 = \biggl( \sum_i r_i \biggr)^2 - 2 \sum_{i\neq j} r_i r_j = n^2 - 2 \cdot \frac{n(n-1)}{2} = n . \] But by the RMS-AM (i.e., Power Mean) and triangle inequalities, \[ 1 = \sqrt{\sum_i r_i^2/n} \ge \sum_i \lvert r_i \rvert / n \ge \biggl\lvert \sum_i r_i/n \biggr\rvert = 1 , \] with equality in the first bound only when $ \lvert r_i \rvert$ is constant, and in the second bound only when all the $ r_i$ have the same sign. Since equality does in fact hold, both of these conditions must be true. It follows that $ r_i = 1$, for all indices $ i$. Thus $ f(x) = (x+1)^n$, and \[ g(x) = (x+1)^n - \left[ x^n + nx^{n-1} + \frac{n(n-1)}{2} x^{n-2} \right] = \sum_{i=0}^{n-3} \binom{n}{i} x^i . \qquad \blacksquare \]

Know, $ P(x)=(x+x_1)(x+x_2)...(x+x_n)$, where $ x_i \in R^+$, $ 1\leq i \leq n$ Then $ \Sigma x_i=n$, $ \Sigma x_ix_j=n(n-1)/2$, $ n^2=(\Sigma x_i)^2=(\Sigma x_i^2) + 2(\Sigma x_ix_j)=(\Sigma x_i^2)+n^2-n$, Then. $ (\Sigma x_i^2)=n$. Then. $ (\Sigma x_i^2)-2(\Sigma x_ix_j)+n=0=\Sigma (x_i-1)^2$->$ x_i=1$, $ 1\leq i \leq n$ Then: $ P(x)=(x+1)^2$, $ g(x)=(x+1)^2-x^n-nx^{n-1}-x^{n-2}n(n-1)/2$

as the polynomial has real roots ( $ r_{i}$ for $ i$ from $ 1$ to $ n$) $ (n - 1) \sum (roots)^{2}$ $ \ge$ $ 2( \sum$ pairwise product of roots) with equality if and only if all roots are equal (as $ rhs - lhs = \sum (r_{i} - r_{j})^{2}$ as $ \sum r_{i}^{2} = ( \sum r_{i} )^{2} - 2 (\sum r_{i}r_{j})$ by Vieta's theorem $ \sum r_{i} = -n$ and $ \sum r_{i}r_{j} = \frac {n(n - 1)}{2}$ thus plugging values we get equality for the inequality mentioned above hence all roots are equal and hence are $ = \frac{-n}{n} = - 1$ $ f(x) = (x + 1)^{n}$ $ g(x) = (x + 1)^{n} - x^{n} - \frac {n(n - 1)}{2} x^{n - 1}$

Let $ a_1,...,a_n$ be the roots of the polynomial $ f(x)$ .Because they are all real we have that $ \sum_{i\neq j} (a_i-a_j)^2\geq 0$ which means that $ (n-1)\sum a_i^2\geq 2\sum_{i\neq j}a_ia_j$ or $ (n-1)(\sum a_i)^2\geq 2n\sum_{i\neq j} a_ia_j$ (1) From Vieta we have that $ \sum a_i=n$ and $ \sum_{i\neq j} a_ia_j=\frac{n(n-1)}{2}$ Now from these two we observe that the equality holds at (1) so all roots are equal so $ f(x)=(x-k)^n$ and equating the coefficient of $ x^{n-1}$ we find that $ k=-1$ so $ f(x)=(x+1)^n$ and so $ g(x)=\sum_{k=3}^n\binom{n}{k}x^{n-k}$

$ p(x)$ must have only real roots and be of the form: \[ p(x) = x^n + nx^{n - 1} + \frac {n(n - 1)}{2} x^{n - 2} + g(x), \] with $ deg (g(x)) \leq n - 3$, and $ g(x)$ is a polynomial. $ p(x)$ has $ n$ not necessarily distinct real roots: $ a_1, a_2, \ldots, a_n$. By Vieta's theorem, it follows \[ a_1 + a_2 + ... + a_n = - n . \] It also follows that $ S_2$, the sum of the product of the roots of $ p(x)$ taken 2 at a time is $ \frac {n(n - 1)}2$. Now consider the following identity: \[ (a_1 + a_2 + \ldots + a_n)^2 = a_1^2 + a_2^2 + \ldots + a_n^2 + 2S_2 \] \[ n^2 = a_1^2 + a_2^2 + \ldots + a_n^2 + n^2 - n \] \[ n = a_1^2 + a_2^2 + \ldots + a_n^2 \] \[ 0 = a_1^2 + a_2^2 + \ldots + a_n^2 + a_1 + a_2 + \ldots + a_n \] Note the average of $ a_1^2, a_2^2,\dots,a_n^2$ is equal to 1, and the average of $ a_1, a_2, \ldots, a_n$ is equal to -1. Algebraically, this translates into \[ \frac {a_1^2 + a_2^2 + \ldots + a_n^2}{n} = - \frac {a_1 + a_2 + \ldots + a_n}n \] \[ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_n^2}{n}} = \sqrt { - \frac {a_1 + a_2 + \ldots + a_n}n} \] Since both sides of the equality are equal to 1, \[{ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_n^2}n} = |\frac {a_1 + a_2 + \ldots + a_n}n|} \] It is well-known that the root mean square (RMS) is greater than or equal to the arithmetic mean (AM) for positive numbers, equality holding if and only if $ a_1 = a_2 = \ldots = a_n$. Although it is not entirely necessary given the restraining conditions of the problem, we will extend the $ RMS\geq AM$ inequality to all real numbers for completeness. To allow for zero terms, without loss of generality we will assume $ a_1 = a_2 = ... = a_j = 0$, and $ a_{j + 1}, a_{j + 2},\ldots, a_n$ are the positive terms. Substituting, since all the zero terms disappear, the RMS of $ a_{j + 1},a_{j + 2},\ldots,a_n$ is greater than or equal to the AM, equality holding when they are all equal. So $ RMS\geq AM$ for nonnegative numbers if and only if all the nonzero terms are equal. Now we will introduce negative numbers. If $ a_1, a_2, \ldots, a_n$ are all nonpositive, then the absolute value sign makes everything positive or zero, and this is equivalent to the nonnegative case. (Without the absolute value condition of the problem, $ RMS \geq AM$ in this case because a positive or zero number (RMS) is always greater than or equal to a zero or negative number (AM), equality holding when they are both zero. For all the following cases, $ RMS > AM$, but for the purposes of this problem we must show $ RMS > |AM|$.) If some elements are nonnegative and others negative, then without loss of generality let $ a_1, a_2, \ldots, a_k$ be nonnegative and $ a_{k + 1}, a_{k + 2}, \ldots, a_n$ be negative. There are three possibilities: (1) $ \displaystyle\sum_{s = 1}^k a_s = \displaystyle\sum_{s = k + 1}^n a_s$ (2) $ \displaystyle\sum_{s = 1}^k a_s > \displaystyle\sum_{s = k + 1}^n a_s$ (3) $ \displaystyle\sum_{s = 1}^k a_s < \displaystyle\sum_{s = k + 1}^n a_s$ If (1), then the AM is 0, and ${ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_n^2}n} > |\frac {a_1 + a_2 + \ldots + a_n}n|}$, and equality clearly never holds. If (2), then after one considers \[{ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_k^2}n} \geq |\frac {a_1 + a_2 + \ldots + a_k}n|}, \] the placement of each of $ a_{k + 1}, a_{k + 2}, \ldots, a_n$ makes the right hand side strictly larger and the left hand side strictly smaller, so \[{ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_n^2}n} > |\frac {a_1 + a_2 + \ldots + a_n}n|}. \] Case (3) is really equivalent to case (2) because of the absolute value sign. More precisely, consider the sequence $ {b_i}$ defined as $ b_i = \{b_i | b_i = - a_i \quad \forall \quad 1\leq i\leq n\}$. Since $ |x| = | - x|$, \[ |\frac {\displaystyle\sum_{i = 0}^n a_i }{n}| = |\frac {\displaystyle\sum_{i = 0}^n b_i }{n}|. \] $ \displaystyle\sum_{i = 1}^k b_i > \displaystyle\sum_{i = k + 1}^n b_i$, so (1) applies on $ b_n$, and ${ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_k^2}n} > |\frac {a_1 + a_2 + \ldots + a_k}n|}.$ We have shown so far that \[{ \sqrt {\frac {a_1^2 + a_2^2 + \ldots + a_k^2}n} \geq |\frac {a_1 + a_2 + \ldots + a_k}n|}. \] for all real numbers $ a_1, a_2, \ldots, a_n$, equality holding if and only if all the nonzero elements of $ \{a_i\}$ are equal. Now let $ a_1, a_2, \ldots, a_u$ be the elements of $ \{a_i\}$ not equal to zero. Also let $ a_{u + 1}, a_{u + 2}, \ldots, a_n$ be the elements of $ \{a_i\}$ equal to zero, if any exist. We have just established $ a_1 = a_2 = \cdots = a_u$ because the equality part of the inequality must hold. Now I will reintroduce the following equation and substitute: \[ a_1^2 + a_2^2 + \ldots + a_n^2 + a_1 + a_2 + \ldots + a_n = 0 \] \[ a_1^2 + a_2^2 + \ldots + a_u^2 + a_1 + a_2 + \ldots + a_u = 0 \] \[ u(a_1)^2 + u(a_1) = 0 \] \[ ua_1(a_1 + 1) = 0 \] Since $ u\neq0$ (if $ u = 0$, then all the roots of the polynomial are 0, so $ p(x) = x^n$, which implies $ n = 0$ because the $ nx^{n - 1}$ must disappear, so $ deg (g(x)) \leq n - 3 = - 3$, which is impossible), and $ a_1 \neq0$, $ a_1 = - 1$. So $ p(x)$ could only be a polynomial with roots of -1 and 0. So, since $ p(x)$ is monic and has only the aforementioned roots, it is well-known that \[ p(x) = (x - 0)^y(x + 1)^z, \] where $ y$ and $ z$ are whole numbers. Expanding, \[ p(x) = x^y(x^z + zx^{z - 1} + \frac {z(z - 1)}2x^{z - 2} + \ldots + 1). \] If $ y\neq 0$, then $ deg(p(x)) = n = y + z$, but the coefficient of $ x^{n - 1}$ is $ z$, and $ z \neq z + y$, reaching contradiction. So $ y \neq 0$, and $ p(x) = (x + 1)^z = (x + 1)^n \quad \forall \quad n \geq 3$. $ n\geq 3$ because $ g(x)$ must have degree $ n - 3$ or less, and the degree of a polynomial cannot be negative. Note this $ p(x)$ has satisfied (and is the only one to satisfy) all the conditions of the problem except the existence of the polynomial $ g(x)$, whose consideration will complete the proof. The definition of a polynomial is an expression of the form $ a_vx^v + a_{v - 1}x^{v - 1} + \ldots + v_0$, where $ v \geq 0$. So $ g(x) = 0$ is a polynomial, and for every $ p(x) = (x + 1)^n$ there is a corresponding polynomial $ g(x)$ given by \[ g(x) \& = (x + 1)^n - x^n - nx^{n - 1} - \frac {n(n - 1)}2 x^{n - 2} \& = n \choose 3 x^{n - 3} + \ldots + 1 \] These $ g(x)$ satisfy all the conditions of the problem and are the only ones to do so; we are done.

Let $ x_1,x_2,...,x_n$ be $ n$ real roots of $ f(x)$.By Viete theorem, we have $ x_1 + x_2 + ... + x_n = - n$. Denote $ y_i = x_i + 1$ ($ i = 1,2,...,n$), then $ y_1 + y_2 + ... + y_n = 0$. We have too $ \sum_{i \neq j}{x_ix_j} = \frac {n(n - 1)}{2}$, deducing $ \sum_{i \neq j}{(y_i - 1)(y_j - 1)} = \frac {n(n - 1)}{2}$, hence $ \sum_{i \neq j}{y_iy_j} = \frac {n(n - 1)}{2} - C_n^2 = 0$. On the other hand, observe that $ y_1^2 + y_2^2 + ... + y_n^2 = (y_1 + y_2 + ... + y_n)^2 - \sum_{i \neq j}{y_iy_j} = 0$, therefore $ y_1 = y_2 = ... = y_n = 0$. Consequently $ x_1 = x_2 = ... = x_n = - 1$. Thus $ f(x) = (x + 1)^n$ and $ g(x) = f(x) - (x^n + nx^{n-1} + \frac{n(n - 1)}{2}x^{n-2}) = \sum_{i = 3}^{n}{C_n^ix^{n - i}}$.