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$ 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.

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    What restrictions on the $x_k$ -- are they to be complex, real, integers, positive integers? – 2017-01-27
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    @coffeemath thanks for pointing out – 2017-01-27
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    With complex numbers, there is a plenty of other solutions. Guess we're talking about real numbers. – 2017-01-27

2 Answers 2

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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 $x_1=...=x_{1994}=1$.

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$$1994\sum_{i=1}^{1994}x^4\geq\sum_{i=1}^{1994}x_i\sum_{i=1}^{1994}x_i^3$$ 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.

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    @Macavity Yes, of course, but for non-negative variables. For all reals we can use the above reasoning. Thank you! – 2017-01-27
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    @Macavity Help me please here: math.stackexchange.com/questions/2113372 . Thank you! – 2017-01-27