Chứng minh rằng : $\large{\sum_{i,j=1}^{n}\tfrac{i.j}{i+j-1}a_{i}a_{j}\geqslant (\sum_{i=1}^{n}a_{i})^2}$

Chứng minh rằng : $$\large{\sum_{i,j=1}^{n}\tfrac{i.j}{i+j-1}a_{i}a_{j}\geqslant (\sum_{i=1}^{n}a_{i})^2}$$
với $a_{i} \in \mathbb{R}$.
Lời giải
Để ý rằng:
$$\int_{0}^{1}x^{i+j-2}dx=\frac{1}{i+j-1}$$
Như vậy:
$$\sum_{i,j=1}^{n}\tfrac{ij}{i+j-1}a_{i}a_{j}=\sum_{i,j=1}^{n}ia_{i}ja_{j}\int_{0}^{1}x^{i+j-2}dx$$
$$=\int_{0}^{1}\left(\sum_{i,j=1}^{n}ia_{i}x^{i-1}.ja_{j}x^{j-1} \right)=\int_{0}^{1}\left(\sum_{k=1}^{n}ka_{k}x^{k-1} \right)^2dx$$
Theo C-S:
$$\int_{0}^{1}\left(\sum_{k=1}^{n}ka_{k}x^{k-1} \right)^2dx \ge \left[\int_{0}^{1}\left(\sum_{k=1}^{n}ka_{k}x^{k-1} \right)dx \right]^2=\left[\sum_{k=1}^{n}ka_{k}\left(\int_{0}^{1}x^{k-1}dx \right) \right]^2=\left(\sum_{k=1}^{n}a_{k} \right)^2$$