Bugi wrote: Let $ a,b,c$ be positive real numbers. Prove the inequality: $ \frac {a^3}{b^2} + \frac {b^3}{c^2} + \frac {c^3}{a^2}\ge \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a}$ By cauchy-swartz: $ (\sum_{cyc} \frac {a^3}{b^2})(\sum_{cyc} a) \ge \left ( \sum_{cyc} \frac{a^2}{b} \right )^2$. So it suffices to prove: $ \sum_{cyc} \frac{a^2}{b} \ge \sum_{cyc} a$. Which is just cauchy-swartz again: $ (\sum_{cyc} \frac{a^2}{b})(\sum_{cyc} b) \ge \left ( \sum_{cyc} a \right )^2$

Or just use this lemma $ \frac{a^3}{b^2}\geq \frac{a^2}{b}+a-b$

Consider the Muirhead's inequality for the real vectors: $ \lambda = (3,0, - 2)$ and $ \mu = (2,0, - 1) \implies (3,0, - 2) \succ (2,0, - 1)$ $ \implies \displaystyle \dfrac {a^3}{b^2} + \displaystyle \dfrac {a^3}{c^2} + \displaystyle \dfrac {b^3}{a^2} + \displaystyle \dfrac {b^3}{c^2} + \displaystyle \dfrac {c^3}{a^2} + \displaystyle \dfrac {c^3}{b^2} \ge \displaystyle \dfrac {a^2}{b} + \displaystyle \dfrac {a^2}{c} + \displaystyle \dfrac {b^2}{a} + \displaystyle \dfrac {b^2}{c} + \displaystyle \dfrac {c^2}{a} + \displaystyle \dfrac {c^2}{b}$ This last expression is the sum of the two following inequalities: $ \displaystyle \dfrac {a^3}{b^2} + \displaystyle \dfrac {b^3}{c^2} + \displaystyle \dfrac {c^3}{a^2} \ge \displaystyle \dfrac {a^2}{b} + \dfrac {b^2}{c} + \displaystyle \dfrac {c^2}{a}$ $ \displaystyle \dfrac {a^3}{c^2} + \displaystyle \dfrac {b^3}{a^2} + \displaystyle \dfrac {c^2}{b} \ge \displaystyle \dfrac {a^2}{c} + \displaystyle \dfrac {b^2}{a} + \displaystyle \dfrac {c^2}{b}$. And we are done.

I think it's not a good solution... You prove that at least one of the inequalities is good.

limes123 wrote: I think it's not a good solution... You prove that at least one of the inequalities is good. Yes, I saw my mistake. I'm going to edit my solution sorry

makmat wrote: Consider the Muirhead's inequality for the real vectors: $ \lambda = (3,0, - 2)$ and $ \mu = (2,0, - 1) \implies (3,0, - 2) \succ (2,0, - 1)$ $ \implies \displaystyle \dfrac {a^3}{b^2} + \displaystyle \dfrac {a^3}{c^2} + \displaystyle \dfrac {b^3}{a^2} + \displaystyle \dfrac {b^3}{c^2} + \displaystyle \dfrac {c^3}{a^2} + \displaystyle \dfrac {c^3}{b^2} \ge \displaystyle \dfrac {a^2}{b} + \displaystyle \dfrac {a^2}{c} + \displaystyle \dfrac {b^2}{a} + \displaystyle \dfrac {b^2}{c} + \displaystyle \dfrac {c^2}{a} + \displaystyle \dfrac {c^2}{b}$ This last expression is the sum of the two following inequalities: $ \displaystyle \dfrac {a^3}{b^2} + \displaystyle \dfrac {b^3}{c^2} + \displaystyle \dfrac {c^3}{a^2} \ge \displaystyle \dfrac {a^2}{b} + \dfrac {b^2}{c} + \displaystyle \dfrac {c^2}{a}$ $ \displaystyle \dfrac {a^3}{c^2} + \displaystyle \dfrac {b^3}{a^2} + \displaystyle \dfrac {c^2}{b} \ge \displaystyle \dfrac {a^2}{c} + \displaystyle \dfrac {b^2}{a} + \displaystyle \dfrac {c^2}{b}$. And we are done. That's a wrong solution. $ a+b \ge 2c$ doesn't nescesarily mean $ a \ge c$ and $ b \ge c$. But you can do it with AM-GM: $ \frac{14\frac{a^3}{b^2}+3\frac{b^3}{c^2}+2\frac{c^3}{a^2}}{19} \ge \frac{a^2}{b}$

I think these inequalities for positive numbers are more interesting: $ (1) \ \ \ \ \ \ \ \ \ \ \ \ \ \frac {a^6}{b^3}+\frac {b^6}{c^3}+\frac {c^6}{a^3}\ge \frac {b^4}{a}+\frac {c^4}{b}+\frac {a^4}{c}$; $ (2) \ \ \ \ \ \frac {a^{14}}{b^7}+\frac {b^{14}}{c^7}+\frac {c^{14}}{d^7}+\frac {d^{14}}{a^7}\ge \frac {b^8}{a}+\frac {c^8}{b}+\frac {d^8}{c}+\frac{a^8}{d}$.

1/$ 3\sum\frac {a^6}{b^3} = \sum(\frac {b^6}{c^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3})\geq 3\sum\frac {b^4}{a}$ 2/$ 7\sum\frac {a^{14}}{b^7} = \sum(4\frac {b^{14}}{c^7} + 2\frac {c^{14}}{d^7} + \frac {d^{14}}{a^7})\geq 7\sum\frac {b^8}{a}$

karis wrote: 1/$ 3\sum\frac {a^6}{b^3} = \sum(\frac {b^6}{c^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3})\geq 3\sum\frac {b^4}{a}$ 2/$ 7\sum\frac {a^{14}}{b^7} = \sum(4\frac {b^{14}}{c^7} + 2\frac {c^{14}}{d^7} + \frac {d^{14}}{a^7})\geq 7\sum\frac {b^8}{a}$ Nice, karis. Can you generalize this inequality to $ n$ positive numbers ?

Vasc wrote: karis wrote: 1/$ 3\sum\frac {a^6}{b^3} = \sum(\frac {b^6}{c^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3})\geq 3\sum\frac {b^4}{a}$ 2/$ 7\sum\frac {a^{14}}{b^7} = \sum(4\frac {b^{14}}{c^7} + 2\frac {c^{14}}{d^7} + \frac {d^{14}}{a^7})\geq 7\sum\frac {b^8}{a}$ Nice, karis. Can you generalize this inequality to $ n$ positive numbers ? We want to show: $ \frac{a^n}{b^{n-k}} + \frac{b^n}{c^{n-k}} + \frac{c^n}{a^{n-k}} \ge \frac{a^m}{b^{m-k}} + \frac{b^m}{c^{m-k}} + \frac{c^m}{a^{m-k}}$ for $ n,m,k \in \mathbb{N}$, $ n > m$, and $ a,b,c > 0$. This can be proven by induction: $ (\sum_{cyc} \frac{a^m}{b^{m-k}})(\sum_{cyc} \frac{a^{m-2k}}{b^{m-3k}}) \ge (\sum_{cyc} \frac{a^{m-k}}{b^{m-2k}})^2$. And it follows easily from here. (Since by using induction $ \sum_{cyc} \frac{a^{m-2k}}{b^{m-3k}} \le \sum_{cyc} \frac{a^{m-k}}{b^{m-2k}}$) (This can be generalized to $ x_1,x_2,..,x_n$ instead of $ a,b,c$ just by changing variables)

Mathias_DK wrote: Vasc wrote: karis wrote: 1/$ 3\sum\frac {a^6}{b^3} = \sum(\frac {b^6}{c^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3})\geq 3\sum\frac {b^4}{a}$ 2/$ 7\sum\frac {a^{14}}{b^7} = \sum(4\frac {b^{14}}{c^7} + 2\frac {c^{14}}{d^7} + \frac {d^{14}}{a^7})\geq 7\sum\frac {b^8}{a}$ Nice, karis. Can you generalize this inequality to $ n$ positive numbers ? We want to show: $ \frac {a^n}{b^{n - k}} + \frac {b^n}{c^{n - k}} + \frac {c^n}{a^{n - k}} \ge \frac {a^m}{b^{m - k}} + \frac {b^m}{c^{m - k}} + \frac {c^m}{a^{m - k}}$ for $ n,m,k \in \mathbb{N}$, $ n > m$, and $ a,b,c > 0$. This is a generalization of the original inequality, not of these inequalities above.

Vasc wrote: Mathias_DK wrote: Vasc wrote: karis wrote: 1/$ 3\sum\frac {a^6}{b^3} = \sum(\frac {b^6}{c^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3})\geq 3\sum\frac {b^4}{a}$ 2/$ 7\sum\frac {a^{14}}{b^7} = \sum(4\frac {b^{14}}{c^7} + 2\frac {c^{14}}{d^7} + \frac {d^{14}}{a^7})\geq 7\sum\frac {b^8}{a}$ Nice, karis. Can you generalize this inequality to $ n$ positive numbers ? We want to show: $ \frac {a^n}{b^{n - k}} + \frac {b^n}{c^{n - k}} + \frac {c^n}{a^{n - k}} \ge \frac {a^m}{b^{m - k}} + \frac {b^m}{c^{m - k}} + \frac {c^m}{a^{m - k}}$ for $ n,m,k \in \mathbb{N}$, $ n > m$, and $ a,b,c > 0$. Vasc wrote: This is a generalization of the original inequality, not of these inequalities above. Could you please explain what the generalisation should be then? (This does also include the inequalities you wrote above, so I'm a bit confused about which generalisation you are talking about ) The first you get by $ n=6>m=4$ and $ k=3$. The second by $ n = 14>m=8$ and $ k=7$.

Mathias_DK wrote: The first you get by $ n = 6 > m = 4$ and $ k = 3$. The second by $ n = 14 > m = 8$ and $ k = 7$. For $ n = 6 > m = 4$ and $ k = 3$, we get $ \frac {a^6}{b^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3}\ge \frac {a^4}{b} + \frac {b^4}{c} + \frac {c^4}{a}$, which is not the desired inequality $ \frac {a^6}{b^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3}\ge \frac {b^4}{a} + \frac {c^4}{b} + \frac {a^4}{c}$.

Vasc wrote: Mathias_DK wrote: The first you get by $ n = 6 > m = 4$ and $ k = 3$. The second by $ n = 14 > m = 8$ and $ k = 7$. For $ n = 6 > m = 4$ and $ k = 3$, we get $ \frac {a^6}{b^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3}\ge \frac {a^4}{b} + \frac {b^4}{c} + \frac {c^4}{a}$, which is not the desired inequality $ \frac {a^6}{b^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3}\ge \frac {b^4}{a} + \frac {c^4}{b} + \frac {a^4}{c}$. You are right! I missed that

Generalize with 3 variable Let $ a;b;c\geq 0: n;m;k\in Z^+;n>m>k$ satisfying $ \frac{n}{m}+\frac{k}{n}=2:$ $ \sum\frac{a^n}{b^{n-k}}\geq\frac{b^m}{a^{m-k}}$

Very nice inequality with Cauchy-Schwarz.

Vasc wrote: karis wrote: 1/$ 3\sum\frac {a^6}{b^3} = \sum(\frac {b^6}{c^3} + \frac {b^6}{c^3} + \frac {c^6}{a^3})\geq 3\sum\frac {b^4}{a}$ 2/$ 7\sum\frac {a^{14}}{b^7} = \sum(4\frac {b^{14}}{c^7} + 2\frac {c^{14}}{d^7} + \frac {d^{14}}{a^7})\geq 7\sum\frac {b^8}{a}$ Nice, karis. Can you generalize this inequality to $ n$ positive numbers ? Straightforward: Let $ m: = 2^{n - 1} - 1 = \sum_{j = 0}^{n - 2}2^{n - 2 - j}$. Then \[ m\sum_{cyc}\frac {a_i^{2m}}{a_{i + 1}^m} = \sum_{cyc}\left(\sum_{j = 0}^{n - 2}2^{n - 2 - j}\frac {a_{j + 1}^{2m}}{a_{j + 2}^{m}}\right)\stackrel{AM - GM }{\ge}m\sum_{cyc}\frac {a_{i + 1}^{m + 1}}{a_{i}}.\] After AM-GM, each inner sum (or rather product then) telescopes.

Inequality Of BMO Let a1,a2,…,an＞0，an＋1= a1，prove that i=1Σnaik＋1aik＋1≥i=1Σnaikai＋1k1, As aik＋1＋ai＋1k＋1≥aikai＋1＋aik＋1ai, i=1,2,…,n. So aik＋1aik＋1＋ai＋1≥aikai＋1k1＋ai. aik＋1aik＋1≥aikai＋1k1＋ai－ai＋1.so i=1Σnaik＋1ika＋1≥i=1Σnaikai＋1k1.

The generalization is very nice inequality if you do not use Muirhead.

Bugi wrote: Let $ a,b,c$ be positive real numbers. Prove the inequality: $$ \frac {a^3}{b^2} + \frac {b^3}{c^2} + \frac {c^3}{a^2}\ge \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a}$$ For $a,b,c>0$ such that $a+b+c=3$. Prove that $$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge 2(a^2+b^2+c^2)-3$$

Bugi wrote: Let $ a,b,c$ be positive real numbers. Prove the inequality: $ \frac {a^3}{b^2} + \frac {b^3}{c^2} + \frac {c^3}{a^2}\ge \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a}$

I'm pretty sure my solution is wrong, thanks to @ below

Greenleaf5002 wrote: Note that the sequences $[a^4c, b^4a, c^4b]$ and $\left[\frac{1}{a^2bc},\frac{1}{ab^2c},\frac{1}{ab^2c}\right]$ are oppositely sorted. Firstly, maybe do you mean $[a^4c, b^4a, c^4b]$ and $\left[\frac{1}{a^2bc},\frac{1}{ab^2c},\frac{1}{abc^2}\right]$? If so, are you sure that $a\geq b\geq c$ gives $a^4c\geq b^4a\geq c^4b$?

redacted.

Greenleaf5002 wrote: as we cannot assume $a\geq b\geq c$ for a cyclic expression anyway But your second triple is symmetric. OK. Explain please why these sequences are oppositely sorted?

What is JBMO ?

Junior Balkan Math Olympiad

Is it possible to use rearrangement inequality?

Clearing the denominators, we have \[a^5c^2+b^5a^2+c^5b^2\geq a^4bc^2+b^4ca^2+c^4ab^2.\]Let $a\geq b \geq c$, then use $\{a^4,b^4,c^4\}$ with $\{ac^2,ba^2, cb^2\}$ and $\{bc^2, ca^2, ab^2\}$ and by rearrangement inequality, the result follows.

Hello! $\sum{\frac{a^3}{b^2}} \geq \sum{\frac{a^2}{b}}$ $\text{With Rearrangement inequality}$ $\implies $ $\boxed{A:\frac{c^2}{a}\geq\frac{b^2} {c}\geq\frac{a^2}{b}}$ $\boxed{B:\frac{c}{a}\geq\frac{b}{c}\geq\frac{a}{b}}$ $3\sum{\frac{a^3}{b^2}}\geq (\sum{\frac{a^2}{b}})(\sum{\frac{a}{b}})\geq 3(\sum{\frac{a^2}{b}})$ $\sum{\frac{a}{b}}\geq3$ Which is true for Am-Gm And we are done

Using Cauchy-Schwarz twice you get the desired result.