This is a question that stumped me during an exam I took today.
Let $c_1,...,c_n,l_1,...,l_n$ be vectors of $\mathbb R^n$ and $\langle .,.\rangle$ denote the dot product.
Prove that $$\sum_{1\leq i,j \leq n} \langle c_i ,c_j \rangle \times \langle l_i ,l_j \rangle \leq \left( \sum_{i=1}^n||c_i||^2 \right)\left( \sum_{i=1}^n||l_i||^2 \right)$$
Expanding a bit it is equivalent to proving that $$\sum_{1\leq i \neq j \leq n} \langle c_i ,c_j \rangle \times \langle l_i ,l_j \rangle \leq \sum_{1\leq i\neq j\leq n} ||c_i||^2 ||l_i||^2 $$
But I don't know how to tackle this one...
After a double application of Schwarz inequality, we obtain $$\langle c_i,c_j\rangle\cdot\langle l_i,l_j\rangle\leqslant \lVert c_i\rVert\cdot \lVert c_j\rVert\cdot\lVert l_i\rVert\cdot \lVert l_j\rVert,$$ hence, after a summation over $i$ and $j$, $$\sum_{1\leqslant i,j\leqslant n}\langle c_i,c_j\rangle\cdot\langle l_i,l_j\rangle\leqslant\left(\sum_{i=1}^n\lVert c_i\rVert\cdot \lVert l_i\rVert\right)^2.$$ We conclude by an other application of Schwarz inequality to the vectors $x:=(\lVert c_i\rVert)_{1\leqslant i\leqslant n}$ and $y:=(\lVert l_i\rVert)_{1\leqslant i\leqslant n}$.