Proof of the nontrivial expression

[sludger13]

I'm looking for a proof of a validity of the inequation:
$(n-1)\sum_{i=1}^{n}x_{i}^{2}\neq 2\sum_{i=1,j=1,j<i}^{n}x_{i}x_{j}$

Assumptions:
$n\geq 2$

$\exists (i,j),x_{i}\neq x_{j}$
$i=1,...,n$
$j=1,...,n$

I have no idea how to prove those non-trivial expressions.

 

[Simon Bridge]

Start by playing around with the expression ... i.e. try it for n=2 and n=3 and see if you can spot a pattern.

There are lots of approaches to proofs that you know.
i.e. induction, assume the converse, and so on.
Have you tried any?

It is very common that there is no obvious way to proceed but you have to have a go anyway.

 

[sludger13]

I started expanding the expression for (n=2,3,4), but it's getting a little chaotic ... I haven't found any pattern so far.
When I try the induction for (n=2) higher, I receive another expression possibly hiding some pattern.

 

[sludger13]

$n=2$:
$x_{1}^{2}+x_{2}^{2}\neq 2x_{1}x_{2}$$n=3$:
$2x_{1}^{2}+2x_{2}^{2}+2x_{3}^{2}\neq 2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}$$n=4$:
$3x_{1}^{2}+3x_{2}^{2}+3x_{3}^{2}+3x_{4}^{2}\neq 2x_{1}x_{2}+2x_{1}x_{3}+2x_{1}x_{4}+2x_{2}x_{3}+2x_{2}x_{4}+2x_{3}x_{4}$

 

[sludger13]

And the in-equation IS VALID - it is the proof of ONE stationary point existence for the function of two variables (for entered assumptions) obtained by the method of least squares. It is easy to imagine that the overall difference of measured values and line values is always minimal for some line directive. I just want to learn to prove it mathematically.

 

[Simon Bridge]

So try to prove the converse. Turn the inequality into an equality - then find the condition that must exist t make the equality work. Since you know the statement is true, the condition will contradict at least one of the assumptions.

 

[sludger13]

Solved out. It wasn't so difficult to find the pattern:
$\sum_{i=1,j=1,j<i}^{n}(x_{i}-x_{j})^{2}\neq0$
That is always valid for the given assumptions.

Thank you for your help :)

 

[Simon Bridge]

Yeah, that what it looks like to me -
Well done :)