1
$\begingroup$

Okay so I have two questions (I think they are pretty simple which was why I put them together), both relating to inequalities that are proving to be challenging. I have learned the AM-GM-HM Inequalities, the Rearrangement Inequality and the Cauchy-Schwarz Inequality.

A. Show that for positive reals $a, b, c$, such that $abc \le 1$ $$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$$

I have combed through this website and Google for solutions, hints anything but have for the most part come up empty handed. I have seen solutions very similar to it, but none have helped. I saw this exact solution:

It is easy to prove that $$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge \frac{a + b + c}{\sqrt[3]{abc}}$$ and since $abc \le 1$then $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$ as required.

It is not clear to me how they arrived at this in particular $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge \frac{a + b + c}{\sqrt[3]{abc}}$. I feel like it is obvious, but I cant see it. So if someone could explain what inequality/trick was used, I would be grateful.

My attempt (which I think is completely wrong) is as follows.

$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$

by AM-GM inequality $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}} = 3$

also by AM-GM $a+b+c \ge 3\sqrt[3]{abc}$

but $abc\le 1$

$\implies 3\sqrt[3]{abc} \le 3$

and thus

$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$

B. Let $a_1, a_2, ... , a_n$ be distinct positive integers. Prove that $$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{1}{1} + \frac{1}{2} +...+ \frac{1}{n}$$

A friend of mine decided to use the Rearrangement Inequality, but I don't really see that right off the bat. I tried to use Cauchy/Schwarz but I didn't get too far. If someone could give a hint or a nudge in the right direction as to which inequality I should use for this one I would appreciate it.

$\endgroup$
1
$\begingroup$

For the first inequality, the trick is to apply the AM-GM inequality to each of the three terms in $${a\over{\sqrt[3]{abc}}} + {b\over{\sqrt[3]{abc}}}+ {c\over{\sqrt[3]{abc}}} .$$ (mouse over for spoiler)

$${a\over{\sqrt[3]{abc}}}=\sqrt[3]{a^3\over abc}=\sqrt[3]{\frac ab\cdot\frac ab\cdot\frac bc}$$

Hint for the second inequality: apply the Rearrangement Inequality to $a_1, a_2, \ldots,a_n$ and ${1\over1^2},{1\over2^2},\ldots,{1\over n^2}$. Once that's done, argue that the result follows.

$\endgroup$
  • $\begingroup$ Thanks very much for your response. I actually got both results today too :) $\endgroup$ – Zhoe Mar 18 '15 at 3:19
0
$\begingroup$

Show that for positive reals, $a, b, c$ such that $abc \le 1$ $$\frac{a}{c} + \frac{c}{b} + \frac{b}{a} \ge a+b+c$$

ANS:

Since $abc \le 1 \implies \frac{1}{abc} \ge 1$ and thus

$$\frac{1}{ac}\ge b, \space \frac{1}{ab}\ge c \space\text{and} \frac{1}{bc}\ge a$$

Now consider $\frac{a}{c} + \frac{a}{c} + \frac{c}{b}$

Using the AM-GM Inequality this gives

$$\frac{a}{c} + \frac{a}{c} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a}{c} \cdot \frac{a}{c} \cdot \frac{c}{b}} = 3\sqrt[3]{\frac{a^2}{bc}}$$

But from above, $\frac{1}{bc} \ge a$

$$\implies \frac{a}{c} + \frac{a}{c} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a^2}{bc}} \ge 3\sqrt[3]{a^3} = 3a$$

Now also consider $\frac{b}{a} + \frac{b}{a} + \frac{a}{c}$ and $\frac{c}{b} + \frac{c}{b} + \frac{b}{a}$ and apply the AM-GM Inequality and use $\frac{1}{ac}\ge b$ and $\frac{1}{ab}\ge c$ from above, we get

$$\implies \frac{b}{a} + \frac{b}{a} + \frac{a}{c} \ge 3\sqrt[3]{\frac{b^2}{ac}} \ge 3\sqrt[3]{b^3} = 3b$$ $$\implies \frac{c}{b} + \frac{c}{b} + \frac{b}{a} \ge 3\sqrt[3]{\frac{c^2}{ab}} \ge 3\sqrt[3]{c^3} = 3c$$

Adding both sides: $$\frac{a}{c} + \frac{a}{c} + \frac{c}{b} +\frac{b}{a} + \frac{b}{a} + \frac{a}{c} +\frac{c}{b} + \frac{c}{b} + \frac{b}{a} \ge 3a+3b+3c \\3\left( \frac{a}{c} +\frac{c}{b}+\frac{b}{a}\right) \ge 3(a+b+c)$$

Let $a_1, a_2, ... , a_n$ be distinct positive integers. Prove that $$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{1}{1} + \frac{1}{2} +...+ \frac{1}{n}$$

ANS:

Let $a_1, a_2,...,a_n$ and $\frac{}{}, \frac{}{},...,\frac{}{}$ be two sequences of reals and $a_1', a_2',...,a_n'$ be a permutation of $a_1, a_2,...,a_n.$ Now since $a_1, a_2,...,a_n$ are distinct positive integers, then $a_i' \ge i \forall i$

Now applying the Rearrangement Inequality we get,

$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{a_1'}{1^2} + \frac{a_2'}{2^2} + ... + \frac{a_n'}{n^2} \ge \frac{1}{1^2} + \frac{2}{2^2} + ... + \frac{n}{n^2} \ge \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} $$

$\endgroup$

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.