Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
Asked
Active
Viewed 204 times
7
-
Hello Stefan, any ideas to resolve. – Jarton Mar 07 '15 at 00:54
-
The right-hand side suggests AM-GM somewhere here, but the inequality goes in the wrong direction for that to work nicely. – Arthur Mar 07 '15 at 00:59
-
Where did you find this inequality? – Stefan4024 Mar 07 '15 at 01:05
-
Are you sure there isn't the RHS isn't multiplied by 2? – Stefan4024 Mar 07 '15 at 01:10
-
Related : https://math.stackexchange.com/questions/313605/prove-that-left1-frac-a-b-right-left1-frac-b-c-right-left1-frac-c – Arnaud D. Jun 04 '20 at 11:18
3 Answers
5
This inequality seems to be somehow too weak, so I'll prove the much stronger inequality:
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$1 + \frac ab + \frac ac \ge 3\sqrt[3]\frac{a^2}{bc} = \frac{3a}{\sqrt[3]{abc}}$$ $$1 + \frac ba + \frac bc \ge 3\sqrt[3]\frac{b^2}{ac} = \frac{3b}{\sqrt[3]{abc}}$$ $$1 + \frac ca + \frac cb \ge 3\sqrt[3]\frac{c^2}{ab} = \frac{3c}{\sqrt[3]{abc}}$$
Sum all the inequalities and we have:
$$3 + \frac ac + \frac ab + \frac ba + \frac bc + \frac ca + \frac cb \ge 3\frac{a+b+c}{\sqrt[3]{abc}}$$
$$1 + \frac ac + \frac ab + \frac ba + \frac bc + \frac ca + \frac cb + 1 \ge 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right) + \left(\frac{a+b+c}{\sqrt[3]{abc}} - 3\right)$$
The last term is obviously positive, since $a+b+c \ge 3\sqrt[3]{abc}$ from AM-GM, hence the proof.
We have equality when $a=b=c$
Stefan4024
- 35,843
-
-
-
@Bernard It's AM-GM on the three numbers $1, \frac{a}{b}, \frac{a}{c}$. – Arthur Mar 07 '15 at 01:05
-
OK. That's what I've just found. For me, it looks like one should use Bernoulli's inequality (and certainly the AM-GM-HM inequalities) to have a more natural calculation. – Bernard Mar 07 '15 at 01:08
3
If the equation out this
$$(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})\geq2(1+\frac{a+b+c}{\sqrt[3]{abc}})$$
We can rewrite the equation:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
writing
$\frac{a+b}{c}=\frac{a+b+c}{c}-1$
$\frac{b+c}{a}=\frac{a+b+c}{a}-1$
$\frac{c+a}{b}=\frac{a+b+c}{b}-1$
Inequality can be written as:
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
Then by AG
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{3(a+b+c)}{\sqrt[3]{abc}}=\frac{2(a+b+c)}{\sqrt[3]{abc}}+\frac{(a+b+c)}{\sqrt[3]{abc}}\geq\frac{2(a+b+c)}{\sqrt[3]{abc}}+3$.
Stefan4024
- 35,843
Angel
- 707
-
-
@martycohen As mentioned in my answer below, the inequality seems somehow weak, since it would also hold if we multiply just the RHS by 2. And I guess Angel is proving a stronger inequality. – Stefan4024 Mar 07 '15 at 01:15
3
The inequality is unchanged if you multiply $a$, $b$, and $c$ by a positive number $k$, so WLOG $abc = 1$. Then the left hand is greater than or equal to
$$1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
and $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$$
since
\begin{align}&\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\\ &= \frac{1}{3}\left(\frac{a}{b} + \frac{a}{b} + \frac{b}{c}\right) + \frac{1}{3}\left(\frac{b}{c} + \frac{b}{c} + \frac{c}{a}\right) + \frac{1}{3}\left(\frac{c}{a} + \frac{c}{a} + \frac{a}{b}\right)\\ &\ge\sqrt[3]{\frac{a}{b}\frac{a}{b}\frac{b}{c}} + \sqrt[3]{\frac{b}{c}\frac{b}{c}\frac{c}{a}} + \sqrt[3]{\frac{c}{a}\frac{c}{a}\frac{a}{b}}\\ &= \frac{a}{\sqrt[3]{abc}} + \frac{b}{\sqrt[3]{abc}} + \frac{c}{\sqrt[3]{abc}}\\ &= a + b + c. \end{align}
kobe
- 41,901