micromass said:
No, it's not that.
Think of a very easy (constant) sequence.
Although this question can be done using the CS Inequality, it can be done this way too:
[tex]a^2+b^2 \geq 2ab[/tex]
[We need to sub into a and b, all the combinations of 2 numbers from the set [tex]\ \ x_1, \ x_2, \ x_2 \ \dots, \ x_n[/tex]
[tex]x_1^2+x_1^2 \geq 2x_1 x_2[/tex]
[tex]x_1^2+x_3^2 \geq 2x_1x_3[/tex]
.
.
.
[tex]x_1^2+x_n^2 \geq 2x_1x_n[/tex]
[tex]x_2^2+x_3^2 \geq 2x_2 x_3[/tex]
And so on, then we add these inequalities side by side.
[tex](n-1)(x_1^2+x_2^2 \dots + x_n^2) \geq 2(x_1x_2+x_1x_3+ \dots)[/tex]
[tex]n(x_1^2+x_2^2+\dots+x_n^2) \geq x_1^2+x_2^2+\dots + x_n^2 + 2(x_1x_2+\dots[/tex]
The RHS is the expanded form of:
[tex]n\sum_{k=1}^{n} x_k^2 \geq \left(\sum_{k=1}^{n}x_k \right)^2[/tex]
[tex]\sum_{k=1}^{n} x_k^2 \geq \frac{1}{n}\left(\sum_{k=1}^{n}x_k \right)^2[/tex]