[carambola]

Demuestra la siguiente igualdad:


$$(\displaystyle\sum_{k=1}^n{a_kb_k})^2 = \displaystyle\sum_{k=1}^n{a_k^2} \displaystyle\sum_{k=1}^n{b_k^2} - \displaystyle\frac{1}{2} \displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^n{(a_ib_j-a_jb_i)^2}$$

[Luis Fuentes]

Hola

Demuestra la siguiente igualdad:


$$(\displaystyle\sum_{k=1}^n{a_kb_k})^2 = \displaystyle\sum_{k=1}^n{a_k^2} \displaystyle\sum_{k=1}^n{b_k^2} - \displaystyle\frac{1}{2} \displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^n{(a_ib_j-a_jb_i)^2}$$

\(  \displaystyle\sum_{k=1}^n{a_k^2} \displaystyle\sum_{k=1}^n{b_k^2} - \displaystyle\frac{1}{2} \displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^n(a_ib_j-a_jb_i)^2=\displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^na_i^2b_j^2- \displaystyle\frac{1}{2} \displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^n(a_ib_j-a_jb_i)^2=\\=\dfrac{1}{2}\displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^n( a_i^2b_j^2+a_j^2b_i^2-(a_ib_j-a_jb_i)^2)=\dfrac{1}{2}\displaystyle\sum_{i=1}^n \displaystyle\sum_{j=1}^n2a_ib_ia_jb_j=\displaystyle\sum_{i=1}^n a_ib_i\displaystyle\sum_{j=1}^na_jb_j=(\displaystyle\sum_{k=1}^na_kb_k)^2 \)

Saludos.