From the condition, we get that there exist $x,y,z\in \mathbb{R}^+$ that $a=\frac{x}{y+z},b=\frac{y}{x+z},c=\frac{z}{x+y}$. So we need to prove $8xyz\leqslant (x+y)(y+z)(z+x)$, which is obvious.

the same thing as this one: http://www.artofproblemsolving.com/community/c6h1152883_prove_that

anantmudgal09 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 $abc \le \frac{1}{8}$. See also here http://www.artofproblemsolving.com/community/c6h1245253p7086270

Full expansion with summation notation quickly gives $ab+bc+ac+2abc=1$ Now just use AM GM on these $4$ terms to get the result.

anantmudgal09 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 $abc \le \frac{1}{8}$. $1= \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ge\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{3+a+b+c}\implies \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\frac{3}{2}\implies abc{\le} \frac{1}{8}.$

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.

Simple amgm

anantmudgal09 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 $abc \le \frac{1}{8}$. http://www.fmat.cl/index.php?showtopic=87045

Is this something to prove??

Why would the permutations work?

how did sqing reach on that step if the the thread is still alive please reply bros and sis(s)

i mean the last step abc < 1/8

Randomizeriitian wrote: i mean the last step abc < 1/8 $ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\frac{3}{2}\implies\frac{3}{2} \ge \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge 3\sqrt[3]{\sqrt{ab}\cdot \sqrt{bc}\cdot \sqrt{ca}}\implies abc{\le} \frac{1}{8}.$

anantmudgal09 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 $abc \le \frac{1}{8}$. Proof

anantmudgal09 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 $abc \le \frac{1}{8}$. Let $n\in\mathbb{N}_{\geq 3}$, $a_i \in \mathbb{R^+}\ \forall 1\leq i\leq n$, such that $\textstyle{\sum_{i=1}^n (1+a_i)^{-1} = n-1}$. Then prove that: \[\sum_{1\leq i < j < k\leq n} \sqrt[3]{a_ia_ja_k} \leq \frac{n(n-2)}{6}.\]

TheOneYouWant wrote: Full expansion with summation notation quickly gives $ab+bc+ac+2abc=1$ Now just use AM GM on these $4$ terms to get the result. Correct

@above you bumped an old topic for nothing. Please don't spam

sqing 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 $$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\frac{3}{2}.$$ Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be positive real numbers such that $\frac{a_1}{1+a_1}+\frac{a_2}{1+a_2}+\cdots+\frac{a_n}{1+a_n}=1.$ Prove that$$\sqrt[n-1]{\frac{T}{a_1}}+\sqrt[n-1]{\frac{T}{a_2}}+\cdots+\sqrt[n-1]{\frac{T}{a_n}}\leq\frac{n}{n-1}.$$Where $T=a_1 a_2 \cdots a_n .$ p/6134379237, p/6114577766

This is not exactly necessary, but i feel its nice. We have $\dfrac{1}{1+a}=\dfrac{b}{1+b}+\dfrac{c}{1+c} \ge 2\sqrt{\dfrac{bc}{(1+b)(1+c)}}$. Multiplying such ineqs, we have the result.

DrYouKnowWho wrote:

Nice work

anantmudgal09 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 $abc \le \frac{1}{8}$. Full expansion of $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$ $\implies \sum_{cyc} ab+2abc=1$ By AM-GM, $\frac{ab+bc+ca+2abc}{4}\geq \sqrt[4]{ab.bc.ca.2abc}=\sqrt[4]{2(abc)^3}$ $\implies 2(abc)^3\leq (\frac 14)^4$ $\implies abc \leq \frac 18$ Hence, it is done.

I think a unique solution (haven't gone through the other posts in the thread); By Titu's lemma, we have: \[ 1 = \frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} \geq \frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{3 + a + b + c} \]which implies \[ 2\sqrt{ab} + 2\sqrt{ac} + 2\sqrt{bc} \leq 3. \]Thus, we have: \[ \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq \frac{3}{2}, \]and since \(\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \geq 3\sqrt[3]{abc}\), it follows that \(\sqrt[3]{abc} \leq \frac{1}{2}\). Cubing both sides yields the result.

Another solution (I guess): $$1- \frac{a}{1+a}+ 1 -\frac{b}{1+b}+1 -\frac{c}{1+c} = 3-1 = \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \Longrightarrow \frac{\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right)}{3} = \frac{2}{3} \geq \frac{3}{2a+2b+2c+3} \Longrightarrow 2a+2b+2c \geq 3 \Longrightarrow \frac{2a+2b+2c}{3} = 1 \geq \sqrt[3]{8abc} \Longrightarrow abc \leq \frac{1}{8}$$