# Given $x_1 + x_2 + .... + x_{1994} = 1994$ find all $x_i$

$x_1 + x_2 + ..... + x_{1994} = 1994$

$x_1^3 + x_2^3 + .... +x_{1994}^3 = x_1^4 + x_2^4 + .... +x_{1994}^4$

Find all $x_i$ where all are real numbers

I tried to prove all are equal to 1 using inequality but couldn't do anything useful. I tried to prove base case where there are only $x_1$ and $x_2$, but no luck that way. Any help would be appreciated.

• What restrictions on the $x_k$ -- are they to be complex, real, integers, positive integers? Commented Jan 27, 2017 at 12:44
• @coffeemath thanks for pointing out Commented Jan 27, 2017 at 12:46
• With complex numbers, there is a plenty of other solutions. Guess we're talking about real numbers. Commented Jan 27, 2017 at 12:47

We can apply the Tchebychev inequality for ordered sequences; it asserts that if for real numbers $a_1,...,a_n$ and $b_1,...,b_n$ we have $a_1≤...≤a_n$ and $b_1≤...≤b_n$ then $$\sum_{i=1}^n a_ib_i≥\frac{1}{n}\left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n b_i\right).$$ with equality if and only if one of the sequences is constant.
We then see that $x_1^3,...,x_{1994}^3$ and $x_1-1,...,x_{1994}-1$ are ordered in the same way and thus $$0=\sum_{i=1}^{1994} x_i^4-\sum_{i=1}^{1994} x_i^3=\sum_{i=1}^{1994} x_i^3(x_i-1)≥\frac{1}{1994}\left(\sum_{i=1}^n x_i^3\right)\left(\sum_{i=1}^n (x_i-1)\right)=0.$$ Hence one of the sequences is constant and therefore $x_1=...=x_{1994}=1$.
$$1994\sum_{i=1}^{1994}x^4\geq\sum_{i=1}^{1994}x_i\sum_{i=1}^{1994}x_i^3$$ The equality occurs for $x_i=1$.
The above inequality it's enough to prove for all $x_i\geq0$, which is true by Muirhead.