Prove $\sum\limits_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}} \geq \frac{(\sum\limits_{i=1}^{n}a_i)^2}{\sum\limits_{i=1}^{n}b_i}$

Question

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

Answer 1

You can simply use the cauchy-scwartz on the sets $$\left\{\frac{a_1}{\sqrt {b_1}},\frac{a_2}{\sqrt {b_2}},\dots,\frac{a_n}{\sqrt {b_n}}\right\}\text{ and }\left\{\sqrt {b_1},\sqrt {b_2},\dots,\sqrt {b_n}\right\}.$$

What you will get, is $$\left(\frac{a_{1}^{2}}{b_{1}} + \frac{a_{2}^{2}}{b_{2}} + \cdot \cdot \cdot +\frac{a_{n}^{2}}{b_{n}}\right)(b_{1}+b_{2}+\dots+b_{n}) \geq (a_{1}+a_{2}+\dots+a_{n})^2.$$

Answer 2

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 $$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$$

Answer 3

From "Mathematical Olympiad Treasures" by Titu Andreescu and Bogdan Enescu

This is further used to prove Cauchy–Schwarz.

Answer 4

Let $A=\sum a_i$, $B=\sum b_i$, with $a_i \in \mathbb{R}$, $b_i \in \mathbb{R}_{>0}$ . Then

$$\sum \frac{a_i^2}{b_i}= B \sum \frac{b_i}{B} \left(\frac{a_i}{b_i}\right)^2= B\, E[c_i^2]$$

where $c_i=\frac{a_i}{b_i}$ is a random variable with probability function $\frac{b_i}{B}$.

But (well known fact, consequence of Jensen inequality) $E[c_i^2] \ge (E[c_i])^2$, with equality iff $c_i$ is constant, i.e., if $b_i$ has only one term. Hence

$$\sum_{i=1}^n \frac{a_i^2}{b_i} \ge B \left( \sum_{i=1}^n \frac{b_i}{B} \frac{a_i}{b_i} \right)^2= \frac{A^2}{B} = \frac{\left( \sum_{i=1}^n a_i\right)^2}{\sum_{i=1}^n b_i}$$ with equality iff $n=1$.