kk108 wrote: Let $a$ , $b$ , and $c$ be positive real numbers with : $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a} $ = 1 . Prove that $abc $ $\le$ 1/8 . Let $abc=w^3$. By condition and AM-GM $1+abc=1+w^3=\sum_{cyc}(a^2c+a^2)\geq3w^3+3w^2$, which gives $w\leq\frac{1}{2}$ or $abc\leq\frac{1}{8}$. Done!

I did something similar actually in RMO .

Something stronger. Let $a$, $b$ and $c$ be positive numbers such that $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that: $$ab+ac+bc\leq\frac{3}{4}$$

arqady wrote: Something stronger. Let $a$, $b$ and $c$ be positive numbers such that $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that: $$ab+ac+bc\leq\frac{3}{4}$$ $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}\ge \frac{(a+b+c)^2}{a+b+c+ab+bc+ca}\ge \frac{(a+b+c)^2}{a+b+c+\frac{(a+b+c)^2}{3}}$ $1\ge \frac{(a+b+c)^2}{a+b+c+\frac{(a+b+c)^2}{3}}\Rightarrow \ \ \ a+b+c\le \frac{3}{2}$. $\frac{9}{4}\ge (a+b+c)^2\ge 3(ab+bc+ca)\Rightarrow \ \ \ ab+bc+ca\le \frac{3}{4}$

Let $a$, $b$ and $c$ be positive numbers such that $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that: $$2(a^2+b^2+c^2)+ab+ac+bc\leq\frac{9}{4}$$

kk108 wrote: Let $a$ , $b$ , and $c$ be positive real numbers with : $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a} $ = 1 . Prove that $abc $ $\le$ 1/8 . RMO Mumbai 2016, P2 $1= \frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}\ge\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{3+b+c+a}\implies \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\frac{3}{2}\implies abc{\le} \frac{1}{8}.$

arqady wrote: Let $a$, $b$ and $c$ be positive numbers such that $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that: $$2(a^2+b^2+c^2)+ab+ac+bc\leq\frac{9}{4}$$

luofangxiang wrote: $a,b,c\geq 0$, such that ${\frac {a}{1+b}}+{\frac {b}{1+c}}+{\frac {c}{1+a}}=1$,prove \[{a}^{2}+{b}^{2}+{c}^{2}+ab+ac+bc\leq 3/2\]

luofangxiang wrote: By the condition and by AM-GM $1+abc=a^2+b^2+c^2+a^2c+b^2a+c^2b\geq a^2+b^2+c^2+3abc$, which gives $a^2+b^2+c^2+2abc\leq1$ and $\{a,b,c\}\subset[0,1]$. Thus, $\sum_{cyc}(a^2+ab)-1=abc+\sum_{cyc}ab(1-b)\geq0$.

This one is easy: $a,b,c\ge 0, \ \ \frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that $$(a+c)^2(2a-1)+(c+b)^2(2c-1)+(b+a)^2(2b-1)\ge 0$$

There exists very simple and short proof for the last inequality.

To know the detailed

luofangxiang wrote: To know the detailed I used only Cauchy-Schwarz.

luofangxiang wrote: To know the detailed I did't understsand you. Do you want full proof? PS. At $ \ (\frac{1}{2}, \frac{1}{2},\frac{1}{2},) \ \ $ we have equality . I have just realized that equality occurs also at $(1,0,0)$

Do you want full proof? Yes, thank you very much

luofangxiang wrote: Do you want full proof? I think some solvers are still trying it. It is better to post full proof later. Is that inequality interesting for you?

这是一个有意思不等式(This is an interesting inequality)

mudok wrote: This one is easy: $a,b,c\ge 0, \ \ \frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that $$(a+c)^2(2a-1)+(c+b)^2(2c-1)+(b+a)^2(2b-1)\ge 0$$ 1.$ \left( 2\,a-1 \right) a+ \left( 2\,b-1 \right) b+ \left( 2\,c-1 \right) c\geq 0 $ 2.$(2a-1)(2a+b)+(2b-1)(2b+c)+(2c-1)(2c+a)\geq 0$

kk108 wrote: Let $a,b,c$ be positive real numbers such that $$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$$Prove that $abc \le \frac{1}{8}$. Substituting $(a=1/x)_{cyc}$, we have $\sum \frac{1}{1+x} =1$. Let $p= \sum x$, $q= \sum xy$, $r=xyz$. Then it follows that $r=2+p$. Since $2+p = 2+x+y+z \ge 4 \sqrt[4]{2r}$, it follows that $r^4 \ge 2^9 r$ i.e. $r \ge 8$ which proves $abc \le \frac{1}{8}$.

kk108 wrote: Let $a,b,c$ be positive real numbers such that $$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$$Prove that $abc \le \frac{1}{8}$. RMO Mumbai 2016, P2

Please one tell whether my proof would be accepted at the RMO- put \(\frac { a }{ 1+a } \) =x and others y,z we obtain x+y+z=1 and a=x/1-x (and cyclic permutations) therefore \( \frac { x }{ 1-x } \frac { y }{ 1-y } \frac { z }{ 1-z } \le \frac { 1 }{ 8 }\) therefore (putting 1=x+y+z) (x+y)(y+z)(z+x)must be greater than 8xyz we can obtain this result by using am-gm on x,y and cyclic permutations and multiplying.

mudok wrote: This one is easy: $a,b,c\ge 0, \ \ \frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that $$(a+c)^2(2a-1)+(c+b)^2(2c-1)+(b+a)^2(2b-1)\ge 0$$ By computer,we have $$LHS=2\sum_{cyc}{\frac{bc(a+b-1)^2}{1+c}}\ge{0}$$

arqady wrote: Let $a$, $b$ and $c$ be positive numbers such that $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that: $$2(a^2+b^2+c^2)+ab+ac+bc\leq\frac{9}{4}$$ By computer,we have $$RHS-LHS=\frac{1}{4}\sum_{cyc}{\frac{b(a+b-2c)^2}{1+c}}+\frac{3}{4}\sum_{cyc}{\frac{ca(a-b)^2}{(1+a)(1+b)}}\ge{0}$$

szl6208 wrote: arqady wrote: Let $a$, $b$ and $c$ be positive numbers such that $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a}=1$. Prove that: $$2(a^2+b^2+c^2)+ab+ac+bc\leq\frac{9}{4}$$ By computer,we have $$RHS-LHS=\frac{1}{4}\sum_{cyc}{\frac{b(a+b-2c)^2}{1+c}}+\frac{3}{4}\sum_{cyc}{\frac{ca(a-b)^2}{(1+a)(1+b)}}\ge{0}$$ You say: conditions is free?

Click to reveal hidden text

$$RHS-LHS-(\frac{1}{4}\sum_{cyc}{\frac{b(a+b-2c)^2}{1+c}}+\frac{3}{4}\sum_{cyc}{\frac{ca(a-b)^2}{(1+a)(1+b)}})$$$$=-\frac{1}{4}\frac{(a^2c+ab^2-abc+bc^2+a^2+b^2+c^2-1)(a^2+5ab+5ac+b^2+5bc+c^2+9a+9b+9c+9)}{(1+c)(1+a)(1+b)}$$

this can be proven by trigonometry too

setting $a=\tan^2{x} ,b=\tan^2{y}, c=\tan^2{z}$ From the identity $\sin^2{A}=\dfrac{\tan^2{A}}{1+\tan^2{A}}$ $\sin^2{x}+\sin^2{y}+\sin^2{z}=1$ $\cos^2{x}=\sin^2{y}+\sin^2{z}$ $\cos^2{y}=\sin^2{x}+\sin^2{z}$ $\cos^2{z}=\sin^2{x}+\sin^2{y}$ By AM-GM $\cos^2{x}\geq 2\sin{y}\sin{z}$ $\cos^2{y}\geq 2\sin{x}\sin{z}$ $\cos^2{z}\geq 2\sin{x}\sin{y}$ Multiplying three inequalities $\cos^2{x}\cos^2{y}\cos^2{z}\geq 8\sin^2{x}\sin^2{y}\sin^2{z}$ $\tan^2{x}\tan^2{y}\tan^2{z}\leq \frac{1}{8}$ $abc\leq \frac{1}{8}$

sqing wrote: kk108 wrote: Let $a$ , $b$ , and $c$ be positive real numbers with : $\frac{a}{1+b} + \frac{b}{1+c} +\frac {c}{1+a} $ = 1 . Prove that $abc $ $\le$ 1/8 . RMO Mumbai 2016, P2 $1= \frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}\ge\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{3+b+c+a}\implies \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\frac{3}{2}\implies abc{\le} \frac{1}{8}.$ Very nice @ sqing sir.

kk108 wrote: Let $a,b,c$ be positive real numbers such that $$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$$Prove that $abc \le \frac{1}{8}$. Here are two solutions from my side

solution 1

solution 2

@below

Take $x = \frac{a}{1+a}$ and so on. So, we have $a = \frac{x}{1-x}$ and so on. The problem becomes- Given positive reals $x,y$ and $z$ such that- $x+y+z =1$, prove that $(1-x)(1-y)(1-z) \ge 8xyz$ This is just AM-GM after you replace $1-x$ by $y+z$ and so on.

My solution is bit easy and I don't know . So we have , $1 =(\sum \frac{a}{1+b})^2$. $\implies 1\ge (a^2+b^2+c^2) (\sum \frac{1}{(1+a)^2})$. [cauchy swertz] $\implies 3(\sum a^2 ) \le \frac{3}{\sum \frac{1}{(a+1)^2} }$. $\implies 3(\sum a^2) \le \sum \frac{(a+1)^2}{3}$.[AM-HM] $\implies 8(\sum a^2 )\le 2(\sum a) +3$. $\implies \frac{8(a+b+c)^2}{3} \le 2(\sum a)+3$. [AM -RMS ] Suppose $w=a+b+c$ . $\implies 8w^2-6w -9\le 0$. $\implies 2w \le 3$. $\implies \frac{w}{3} \le \frac{1}{2}$. Finally by AM-GM we have , $\boxed{abc \le \frac{1}{8}}$.

Let $a,b,c$ be positive real numbers such that $\frac{a}{1+a}+\frac{2b}{1+b}+\frac{c}{1+c}=1.$ Prove that $$abc \le \frac{2\sqrt 3-3}{9}$$Let $a,b,c$ be positive real numbers such that $\frac{a}{1+a}+\frac{2b}{1+b}+\frac{2c}{1+c}=1.$ Prove that $$abc \le \frac{1}{48}$$Let $a,b,c$ be positive real numbers such that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$ Prove that$$ab^2c^2 \le \frac{27}{512}$$

Let $a,b,c$ be positive real numbers such that $\frac{a^2}{1+bc}+ \frac{b^2}{1+ca}+ \frac{c^2}{1+ab}=1.$ Prove that $$a^2+b^2+c^2\leq\frac{3}{2} $$