wait is this contest not ongoing what

Day 2 is over so discussion is allowed

Discussion is allowed - see here. Anyways, does anyone have a solution?

ghu2024 wrote: $$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a}\ge\frac{a}{b}+\frac{b}{c}+\frac{a}{c}$$ isnt it $\frac{c}{a} $?

Yes the problem was : If $a > 1, b > 1, c>1$ then prove that : $$\frac{a^a}{b^b} + \frac{b^b}{c^c} + \frac{c^c}{a^a} \ge \frac{a}{b} + \frac{b}{c} + \frac{c}{a}.$$ This problem was proposed by me, so I definitely have a solution .

Any hint?

I will post the official solution after some hours (After lunch).

chrono223 wrote: Any hint? Maybe proving that $\frac{a^a}{b^b} + \frac{b^b}{a^a}\geq \frac{a}{b} + \frac{b}{a}$ helps, I don't know what else we can do.

Nonameyet wrote: Maybe proving that $\frac{a^a}{b^b} + \frac{b^b}{a^a}\geq \frac{a}{b} + \frac{b}{a}$ helps, I don't know what else we can do. please tell me if its wrong WLOG. $a\ge b$ $f(x)=x+\frac{1}{x} (x\ge 1)$ is minimal when x is close to 1. as $\frac{a^a}{b^b}\ge \frac{a^b}{b^b}\ge \frac{a}{b}$ (because $a,b\ge 1$) $f(\frac{a^a}{b^b})\ge f(\frac{a}{b})$

Solution: Multiplying by $a^ab^bc^c$, it is equivalent to proving that $a^{2a}c^c+b^{2b}a^a+c^{2c}b^b\geq a^{a+1}b^{b-1}c^c+b^{b+1}c^{c-1}a^a+c^{c+1}a^{a-1}b^b$ $$\Leftrightarrow a^{2a}c^c+b^{2b}a^a+c^{2c}b^b-(a^{a+1}b^{b-1}c^c+b^{b+1}c^{c-1}a^a+c^{c+1}a^{a-1}b^b)=a^{a+1}c^c(a^{a-1}-b^{b-1})+b^{b+1}a^a(b^{b-1}-c^{c-1})+c^{c+1}b^b(c^{c-1}-a^{a-1})=a^{a+1}c^c(a^{a-1}-b^{b-1}+b^{b-1}-c^{c-1})+(b^{b+1}a^a-a^{a+1}c^c)(b^{b-1}-c^{c-1})+c^{c+1}b^b(c^{c-1}-a^{a-1})\geq 0$$$$\Leftrightarrow (a^{a-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b)+(b^{b-1}-c^{c-1})(b^{b+1}a^a-a^{a+1}c^c) = (a^{a-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b)-(b^{b-1}-c^{c-1})(a^{a+1}c^c-b^{b+1}a^a)\geq 0$$. Note that the derivative of $x^{x-1}=e^{(x-1)lnx}$ is $e^{(x-1)lnx}\cdot(\frac{x-1}{x}+lnx)\geq 0$ for $x\geq 1$, so for $x,y> 1, x\geq y, x^{x-1}\geq y^{y-1}$. The original inequality is invariant under cycling $a, b, c$ so WLOG, $min (a,b,c)=c \Rightarrow a^{a-1}\geq c^{c-1}, b^{b-1}\geq c^{c-1}$ Case 1: $a\geq b$ Then $a^{a+1}c^c\geq b^{b+1}c^c\geq c^{c+1}b^b \Rightarrow (a^{a-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b) \geq 0 $. If $a^{a+1}c^c-b^{b+1}a^a \leq 0$, then clearly $(a^{a-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b)-(b^{b-1}-c^{c-1})(a^{a+1}c^c-b^{b+1}a^a)\geq 0$. Assume $a^{a+1}c^c-b^{b+1}a^a \geq 0$. Then as $c^{c+1}\leq ba^a , (c^{c+1} \leq c^{a+1} \leq b^{a+1} = bb^a \leq ba^a), c^{c+1}b^b\leq b^{b+1}a^a \Rightarrow a^{a+1}c^c-c^{c+1}b^b\geq a^{a+1}c^c-b^{b+1}a^a \geq 0$. We have $(a^{a-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b)-(b^{b-1}-c^{c-1})(a^{a+1}c^c-b^{b+1}a^a)\geq (a^{a-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b)-(b^{b-1}-c^{c-1})(a^{a+1}c^c-c^{c+1}b^b)$ $$=(a^{a-1}-b^{b-1})(a^{a+1}c^c-c^{c+1}b^b)\geq 0$$as we know $a^{a-1}\geq b^{b-1}$ and $a^{a+1}\geq cb^b\Rightarrow a^{a+1}c^c\geq c^{c+1}b^b$. The proof for $a\leq b$ is very similar, so we are done.

Inequalities

Nonameyet wrote: chrono223 wrote: Any hint? Maybe proving that $\frac{a^a}{b^b} + \frac{b^b}{a^a}\geq \frac{a}{b} + \frac{b}{a}$ helps, I don't know what else we can do. This is doable using Rearrangement. But this doesn't help. There is a very nice trick, which was devised by me and prof. Fedor Petrov, long ago, and based on that I created this problem. I won't upload its solution until tomorrow, because some others want to try it themselves.

Here is my solution in the contest: Let $A=a^{a-1}, B=b^{b-1}, C=c^{c-1}, x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}.$ then obviously $A,B,C$ and $a,b,c$ have the same order. we start now be applying wieghted AM-GM: $LHS=x\cdot \frac{A}{B}+y\cdot \frac{B}{C}+z\cdot \frac{C}{A}\geq (x+y+z)(A^{x-z}B^{y-x}C^{z-y})^{\frac{1}{x+y+z}}$ and as $RHS=x+y+z$, it is enough to verify that: $A^{x-z}B^{y-x}C^{z-y}\geq 1\;\Leftrightarrow x\log A+y\log B+z\log C\geq z\log A+x\log B+y\log C$ Now, assume wlog $a=\max (a,b,c),$ then $x\geq 1\geq z.$ we consider two cases, case 1: $b\geq c,$ then $x,y\geq 1\geq z,$ and by rearrangement we get: $y\log B+z\log C\geq z\log B+y\log C$ , and $x\log A+z\log B\geq z\log A+x\log B$ and the result immediately follows. case 2: $c\geq b,$ then $x\geq 1\geq y,z,$ and by rearrangement we get: $x\log A+z\log C\geq z\log A+x\log C,$ and $y\log B+x\log C\geq x\log B+y\log C$ and again, the result immediately follows. $\square .$

Since $f(x)=x^{x-1}$ is increasing on $(1,+\infty)$, it's clear that $(\frac{a}{b}-1)(\frac{a^{a-1}}{b^{b-1}}-1) \ge 0$. Thus $$LHS-RHS= \sum_{cyc} \frac{a}{b}(\frac{a^{a-1}}{b^{b-1}}-1) \ge \sum_{cyc} (\frac{a^{a-1}}{b^{b-1}}-1) \ge 0$$where the last inequality holds by simply AM-GM. Edit: The idea would be natural if you've seen this.

Rickyminer wrote: Since $f(x)=x^{x-1}$ is increasing on $(1,+\infty)$, it's clear that $(\frac{a}{b}-1)(\frac{a^{a-1}}{b^{b-1}}-1) \ge 0$. Thus $$LHS-RHS= \sum_{cyc} \frac{a}{b}(\frac{a^{a-1}}{b^{b-1}}-1) \ge \sum_{cyc} (\frac{a^{a-1}}{b^{b-1}}-1) \ge 0$$where the last inequality holds by simply AM-GM. Edit: The idea would be natural if you've seen this. This idea gives us an inequality template. Let \(f(x) > 0\) be an increasing function on positive real numbers. Then we have: \[\frac{xf(x)}{yf(y)} + \frac{yf(y)}{zf(z)} + \frac{zf(z)}{xf(x)} \geq \frac{x}{y} + \frac{y}{z} + \frac{z}{x}\]

ghu2024 wrote: If $a, b, c$ are real numbers satisfying $a > 1, b > 1, c > 1$, then prove that: $$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$ Proposed by Aditya Guha Roy

let $f(x)=x^x$. Note $\dfrac{f(x)}{x}$ is an increasing function in $(1,\infty)$, so by rearrangement, $$b\dfrac{f(a)}{a}+a\dfrac{f(b)}{b} \le f(a)+f(b)$$which implies the inequality $$\dfrac{f(a)}{f(b)}-\dfrac{a}{b}\ge \dfrac{bf(a)}{af(b)}-1$$summing up and applying AM GM, we are done