4
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I would say that you should instead use Jensen inequality. $\endgroup$ – Thomas Boulier Jun 12 '16 at 9:51
  • 1
    $\begingroup$ multiply both sides by denominator of right hand side, write $b_i=(\sqrt{b_i})^2$, then stare at the inequality for ten more seconds. $\endgroup$ – gamma Jun 12 '16 at 9:52
  • 2
    $\begingroup$ This can be done by mathematical induction. First show it for n=2 then it can be generalized. This is known as $T_2$'s lemma. $\endgroup$ – zxcvber Jun 12 '16 at 9:53
7
$\begingroup$

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 $$\left(\frac{a_{1}^{2}}{b_{1}} + \frac{a_{2}^{2}}{b_{2}} + \cdot \cdot \cdot +\frac{a_{n}^{2}}{b_{n}}\right)(b_{1}+b_{2}+\dots+b_{n}) \geq (a_{1}+a_{2}+\dots+a_{n})^2.$$

$\endgroup$
4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

enter image description here

This is further used to prove Cauchy–Schwarz.

$\endgroup$
0
$\begingroup$

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 $n=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.