Let
![E_n=(a_1-a_2)(a_1-a_3)\ldots(a_1-a_n)+(a_2-a_1)(a_2-a_3)\ldots(a_2-a_n)+\ldots+(a_n-a_1)(a_n-a_2)\ldots(a_n-a_{n-1}).](/media/m/6/7/a/67a7124b59ea9299879de3d6a3d2b0d7.png)
Let
![S_n](/media/m/c/2/e/c2e97c6b8d807599d8398c14e4ac95fc.png)
be the proposition that
![E_n\ge0](/media/m/4/c/4/4c411e6cc6cadd53df2afb9f122b6c87.png)
for all real
![a_i](/media/m/2/a/2/2a22407e8a19d2df9d425caa379f34a8.png)
. Prove that
![S_n](/media/m/c/2/e/c2e97c6b8d807599d8398c14e4ac95fc.png)
is true for
![n=3](/media/m/6/e/8/6e8cc663572ec564892ed13a28debcb1.png)
and
![5](/media/m/e/a/3/ea36c795dac330f34d395d8364d379b6.png)
, but for no other
![n>2](/media/m/b/8/a/b8a6d2bba9f17a4b18791eda0f2c0bf7.png)
.
%V0
Let $$E_n=(a_1-a_2)(a_1-a_3)\ldots(a_1-a_n)+(a_2-a_1)(a_2-a_3)\ldots(a_2-a_n)+\ldots+(a_n-a_1)(a_n-a_2)\ldots(a_n-a_{n-1}).$$ Let $S_n$ be the proposition that $E_n\ge0$ for all real $a_i$. Prove that $S_n$ is true for $n=3$ and $5$, but for no other $n>2$.