<s>[INST] Prove $\sum\limits_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}} \geq \frac{(\sum\limits_{i=1}^{n}a_i)^2}{\sum\limits_{i=1}^{n}b_i}$ So I have the following problem, which I'm having trouble solving:
Let $a_1$ , $a_2$ , ... , $a_n$ be real numbers. Let $b_1$ , $b_2$ , ... , $b_n$ be positive real numbers. Prove 
$$ \frac{a_{1}^{2}}{b_{1}} + \frac{a_{2}^{2}}{b_{2}} + \cdot \cdot \cdot +\frac{a_{n}^{2}}{b_{n}} \geq \frac{(a_{1}+a_{2}+\cdot \cdot \cdot+a_{n})^2}{b_{1}+b_{2}+\cdot \cdot \cdot+b_{n}} $$
I was thinking that I somehow could use the Cauchy–Schwarz inequality, but with no success.
Any help would be very appreciated
 [/INST] We will use induction.
First, suppose that $n=2$, the given inequality is 
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\geq \frac{(a_1+a_2)^2}{b_1+b_2}$$
This can be proven directly by subtracting, then factoring the equation into a complete square.
Suppose the given inequality holds for $n=k$
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_1}+\cdots+\frac{a_k^2}{b_k}\geq \frac{(a_1+a_2+\cdots+a_k)^2}{b_1+b_2+\cdots+b_k}$$
Add $\frac{a_{k+1}^2}{b_{k+1}}$ on both sides,
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_1}+\cdots+\frac{a_k^2}{b_k}+\frac{a_{k+1}^2}{b_{k+1}}\geq \frac{(a_1+a_2+\cdots+a_k)^2}{b_1+b_2+\cdots+b_k}+\frac{a_{k+1}^2}{b_{k+1}}$$
$$\geq \frac{(a_1+a_2+\cdots+a_k+a_{k+1})^2}{b_1+b_2+\cdots+b_k+b_{k+1}}$$
Last inequality holds from the case $n=2$. Thus the given inequality is true for $n=k+1$.
We are done!
Equality holds when
$$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$$
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