# Prove, that $(\sum_{i=1}^{n}x_i)^3\geq\frac{9n}{2}*[(\sum_{i=1}^{n}x_i)^2-\sum_{i=1}^{n}x_i^2]$

Let n be a natural number, $\displaystyle n\geq2$ and the real non-negative numbers $x_1, x_2,...,x_n$ with the property that $x_1^2+x_2^2+...+x_n^2=3n^2$ Is this inequality true? $\displaystyle (\sum_{i=1}^{n}x_i)^3\geq\frac{9n}{2}*[(\sum_{i=1}^{n}x_i)^2-\sum_{i=1}^{n}x_i^2]$ I asked yesterday a similar question, but I have reasons to think that this is true. Thank you!

Consider function $$f(y)= y^3-\frac{9n}2(y^2-3n^2)$$ Its derivative is $$f'(y)=3y(y-3n)$$ so, $\forall y>0$ $$f(y)\geq f(3n)=0$$ Now take $$y=\sum_{i=1}^{n}x_i$$
• why is $f(y) \geq f(3n)?$ – user85046 Nov 3 '13 at 21:52
• $y>3n \to f'>0 \to f(y)>f(3n). y<3n \to f'<0 \to f(y)>f(3n)$ – chenbai Nov 4 '13 at 3:30