TITLE: Show that if $\prod\limits_{k=1}^{n}(x+a_k)=\sum\limits_{k=0}^{n} {n\choose k}a^k_kx^{n-k}$ then $a_1=a_2=a_3=....=a_{n-1}=a_n$
QUESTION [12 upvotes]: Let $a_0=1$. Prove that, if 
  $$\prod_{k=1}^{n}(x+a_k)=\sum_{k=0}^{n} {n\choose k}a^k_kx^{n-k}=x^n+{n\choose 1}a_1x^{n-1}+{n\choose 2}a^2_2x^{n-2}+....+a^n_n,$$
  then $a_1=a_2=a_3=....=a_{n-1}=a_n$.

I proved for $n=2,3$ but I do not know how to prove for general solution. Any help for general solution will be appreciated .

Solution for $n=2$
$a_0=1$
$$2a_1=a_{1}+a_{2}$$
$$a^2_2=a_{1}a_{2}$$
$$a_2=a_{1}$$

Solution for $n=3$
$a_0=1$
$$3a_1=a_1+a_2+a_3$$
$$3a^2_2=a_1a_2+a_1a_3+a_2a_3$$
$$a^3_3=a_1a_2a_3$$
.
.
$$2a_1=a_2+a_3 \tag1$$
$$3a^2_2=a_1a_2+a_1a_3+a_2a_3 \tag2$$
$$a^2_3=a_1a_2 \tag3$$
.
$$3a^2_2=a_1(a_2+a_3)+a_2a_3=2a^2_1+a_2a_3$$
$$6a^2_2=4a^2_1+2a_2a_3=(a_2+a_3)^2+2a_2a_3$$
$$5a^2_2-4a_2a_3-a^2_3=0$$
$$(5a_2+a_3)(a_2-a_3)=0$$
If we select $a_3=-5a_2$ then
$25a_2=a_1$
To check Equation 1
$$2a_1=a_2+a_3$$
$$50a_2 \neq a_2-5a_2$$
Thus $$(5a_2+a_3)(a_2-a_3)=0$$
We must use 
$a_2-a_3=0$
$a_2=a_3$
To check Equation 1,3
Equation 1:
$$2a_1=a_2+a_3$$  
$$a_1=a_3$$
Equation 3:
$a^2_3=a_1a_2 $
$$a_1=a_3$$
Thus 
$$a_1=a_2=a_3$$

REPLY [6 votes]: The triangular inequality, nothing but (the equality case of) the triangular inequality...

Here are some notations. Fix some complex numbers $(a_i)_{1\leqslant i\leqslant n}$ such that the hypothesis holds. Let $A=\max\{|a_i|\,;\,1\leqslant i\leqslant n\}$. Choose some $1\leqslant K\leqslant n$ such that $|a_K|=A$. If $A=0$ then $a_i=0$ for every $i$ hence the result holds. We assume without loss of generality that $A\gt0$. Yet another notation: for every $I\subseteq\{1,2,\ldots,n\}$, let $a(I)=\prod\limits_{i\in I}a_i$, hence $|a(I)|\leqslant A^{K}$ for every $I$ of size $K$.
The hypothesis says that
$$
{n\choose K}a_K^K=\sum_{|I|=K}a(I).
$$
By the triangular inequality,
$$
{n\choose K}A^K={n\choose K}|a_K^K|=\left|\sum_{|I|=K}a(I)\right|\leqslant\sum_{|I|=K}|a(I)|\leqslant{n\choose K}A^K.
$$
Thus, both inequalities above are in fact equalities. 

Since the first inequality is an equality, there exists $z$ such that $|z|=1$ and $a(I)=z|a(I)|$ for every $I$ such that $|I|=K$. 
Since the second inequality is an equality, $|a(I)|=A^K$ for every $I$ such that $|I|=K$. 

Since $|a_i|\leqslant A$ for every $i$, item 2. yields $|a_i|=A$ for every $i$ (remember that $K\geqslant1$ hence every $i$ belongs to some $I$ such that $|I|=K$). This shows that every $1\leqslant K\leqslant n$ fulfills the hypothesis that $|a_K|=A$, in particular the reasoning above holds for $K=1$. Using item 1. for $K=1$ yields $a_i=zA$ for every $1\leqslant i\leqslant n$, QED.
Edit: This proves the result for $(a_i)_{1\leqslant i\leqslant n}$ in any normed algebra such that the equality case used in the triangular inequality holds, replacing everywhere the modulus $|\ |$ by the norm on the algebra.