For real numbers $a_1,\ldots,a_n,b_1,\ldots,b_n,c_1,\ldots,c_n$ it holds that $$ \sum_{k=1}^n a_kb_kc_k\le \left(\sum_{k=1}^n a_k^2 \right)^{1/2} \left(\sum_{k=1}^n b_k^2 \right)^{1/2} \left(\sum_{k=1}^n c_k^2 \right)^{1/2} . $$
Is there a geometric interpretation for this inequality?
The RHS is simply the product of the Euclidean lengths of the vectors $a$, $b$ and $c$, where $a=(a_1,\ldots,a_n),$ etc. But I'm not sure about the LHS. Does the LHS have a geometric meaning?
I'm thinking that there may be a nice geometric interpretation because this inequality looks a lot like the Cauchy-Schwarz inequality, but with three vectors instead of two.
Here's a proof of this inequality:
If one of the factors on the RHS is zero, the inequality holds trivially. Otherwise divide by the RHS and write $\hat a = {a \over|| a||}$, $\hat b = {b \over|| a||}$, $\hat c = {c \over|| c||}$. Then $$ \sum_{k=1}^n \hat a_k \hat b_k \hat c_k \le \left(\sum_{k=1}^n \hat a_k^2 \right)^{1/2} \left(\sum_{k=1}^n \hat b_k^2 \hat c_k^2 \right)^{1/2} \le \left(\sum_{k=1}^n \hat a_k^2 \right)^{1/2} \left(\sum_{k=1}^n \hat b_k^2 \right)^{1/2} = 1, $$
where the first inequality is due to Cauchy-Schwarz and the second inequality follows because all components of the normed vector $\hat c$ must have absolute value $\le 1$. Multiplying by $||a||\, ||b||\, ||c||$ then yields the claim.