TITLE: Inequality proof by Cauchy-Schwarz inequality
QUESTION [5 upvotes]: I've been wrestling with the following inequality for 3 days to prove it by CSI, but I can't choose the right column vectors $u$ and $v$ of CSI.
$$
\frac{a^2_1}{b_1}+\frac{a^2_2}{b_2}+\cdots+\frac{a^2_n}{b_n}\ge\frac{\left(a_1+\cdots+a_n\right)^2}{b_1+\cdots+b_n}
$$
How on earth should I fix these vectors?

REPLY [4 votes]: As a separate answer from the one I just posted. This time put $x_i = \dfrac{a_i}{b_i}> 0$, and asume: $a_1+a_2+\cdots + a_n = 1 = b_1+b_2+\cdots + b_n$, then we prove: $b_1x_1^2+b_2x_2^2+\cdots + b_nx_n^2\ge 1$. The function $f(x) = x^2$ is convex on $(0,\infty)$, thus Jensen inequality gives: $b_1f(x_1)+b_2f(x_2)+\cdots + b_nf(x_n) \ge f(b_1x_1+b_2x_2+\cdots + b_nx_n)$ translates into $b_1x_1^2+b_2x_2^2+\cdots +b_nx_n^2 \ge f(1) = 1$. 
Done !