it should be $2(a+b+c)+1\ge32abc$, I think (try $a=b=c=\frac12$)

Sorry you have right. I wrote 64 instead of 32 . Try now the problem

OK let's try this: let $\sqrt{ab} = \cos C$ etc so we have $\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C = 1$. Thus, by a fairly well known identity, $A,B,C$ are the angles of a triangle. We have $2(a+b+c) + 1 \geq 32abc$ $\iff 2\sum_{\text{cyc}} \frac{\cos B\cos C}{\cos A} + 1 \geq 32\cos A\cos B\cos C$ By AM-GM, $2\sum_{\text{cyc}} \frac{\cos B\cos C}{\cos A} + 1 \geq 4\sqrt[4]{ 8\cos A\cos B\cos C}$ $\geq 32 \cos A\cos B\cos C$ where our last inequality is equivalent to the well-known $\cos A\cos B\cos C \leq \frac 18$.

silouan wrote: Let $a,b,c>0$ and $ab+bc+ca+2abc=1$ then prove that $2(a+b+c)+1\geq 32abc$ Nice inspiration, silouan! I think substitution $a=\frac{x}{y+z}$ $(x,y,z>0)$ etc as well as AM-GM are enough,aren't they??

ThAzN1 I recalled your nice sub, too! Nice solution!

i'm sorry silouan, but this inequality is not yours...for instance, you can check http://www.campus-oei.org/oim/revistaoim/numero14/CMM20041.pdf my solution consisted in proving $2(a+b+c)+1\geq 4\geq 32abc$, but a friend of mine find the inequality is equivalent to $\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}+\frac{1}{2abc}\geq 16$, which comes immediately from AM-HM or Cauchy-Schwarz

What a pity.... I thought that this problem I constucted was not well-known My solution is the same as your friend solution. But could you post your solution ?As I saw you prove a more stronger ineq

I think stronger inequality holds: $a+b+c+4abc\geq 2$

when i did this i didn't know too much about substitution and that kind of nice tricks, so don't expect to see a nice solution from am-gm on the condition we have that $1\geq 4\sqrt[4]{2(abc)^3} \Rightarrow 4\geq 32abc$ for the other part, from the condition we get $a=\frac{1-bc}{b+c+2bc}$, then substitute $a$ for $\frac{1-bc}{b+c+2bc}$ on what we want to prove, which is clearly equivalent with $2(a+b+c)\geq 3$ then, we would have to show that $2(b+c)(b+c+2bc)+2(1-bc)\geq 3(b+c+2bc)$ or $b^2(4c+2)+b(4c^2-4c-3)+(2c^2-3c+2)\geq 0$, but the discriminant equals $\Delta=(2c+1)^2(2c-3)^2-8(2c+1)(2c^2-3c+2)=(2c+1)(2c-1)^2(2c-7)$, but if $2c-7\geq 0$ then the inequality turns obvious, so we must have $\Delta\leq 0$, and we're done.

fuzzylogic wrote: I think stronger inequality holds: $a+b+c+4abc\geq 2$ Very nice problem, fuzzylogic!

fuzzylogic wrote: I think stronger inequality holds: $a+b+c+4abc\geq 2$ making the substitutions : $a=\frac{x}{y+z}$ ; $b=\frac{y}{x+z}$; $c=\frac{z}{x+y}$ that verify $ab+bc+ca+2abc=1$ and replacing them in $a+b+c+4abc\geq 2$ we obtain after expanding schur

Excuse me, following is my solution. silouan wrote: Let $a,b,c>0$ and $ab+bc+ca+2abc=1$ then prove that $2(a+b+c)+1\geq 32abc$ From the thesis, we have $1=ab+bc+ca+2bac \geq 4\sqrt[4]{2a^{3}b^{3}c^{3}}\Leftrightarrow abc \leq{1 \over 8}$, or $1 \geq 8abc\ (1)$. We also have $ab+bc+ca=1-2abc \geq{3 \over 4}$. $2(a+b+c) \geq 3 \geq 24abc \ (2) \\ \Leftrightarrow a+b+c \geq{3 \over 2}\\ \Leftrightarrow (a+b+c)^{2}\geq{9 \over 4}(ab+bc+ca+2abc) \\ \Leftrightarrow (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+{3 \over 2}\left(ab+bc+ca-{3 \over 4}\right)+9\left({1 \over 8}-abc\right) \geq 0.$ So that, (2) is proved. Adding (1) and (2) up, we get: \[2(a+b+c)+1 \geq 32abc. \]

Could anyone mean AM-GM please? Thanks!

http://www.math10.com/forumbg/viewtopic.php?t=748&start=90 See problem 37.

Strengthen of MMO 2004

fuzzylogic wrote: I think stronger inequality holds: $a+b+c+4abc\geq 2$ Now we can prove it by $uvw$.

arqady wrote: fuzzylogic wrote: I think stronger inequality holds: $a+b+c+4abc\geq 2$ Now we can prove it by $uvw$. Strengthen of MMO 2004 (2)

arqady wrote: fuzzylogic wrote: I think stronger inequality holds: $a+b+c+4abc\geq 2$ Now we can prove it by $uvw$. _________________ The problem is equivalent to: $ a,b,c>0,ab+bc+ca+abc=4\Rightarrow a+b+c+abc\ge 4\Leftrightarrow $ $ a,b,c>0,\frac{a}{2+a}+\frac{b}{2+b}+\frac{c}{2+c}=1\Rightarrow a+b+c\ge ab+bc+ca $ But by C-S we have: $ \frac{a}{2+a}+\frac{b}{2+b}+\frac{c}{2+c}=1\Leftrightarrow \frac{a^2}{2a+a^2}+\frac{b^2}{2b+b^2}+\frac{c^2}{2c+c^2}=1 $ $ \Leftrightarrow 1\ge\frac{(a+b+c)^2}{2(a+b+c)+a^2+b^2+c^2}\Leftrightarrow $ $ \Leftrightarrow a+b+c\ge ab+bc+ca\Leftrightarrow QED $ ____ Sandu Marin

sandumarin wrote: arqady wrote: fuzzylogic wrote: I think stronger inequality holds: $a+b+c+4abc\geq 2$ Now we can prove it by $uvw$. _________________ The problem is equivalent to: $ a,b,c>0,ab+bc+ca+abc=4\Rightarrow a+b+c+abc\ge 4\Leftrightarrow $ $ a,b,c>0,\frac{a}{2+a}+\frac{b}{2+b}+\frac{c}{2+c}=1\Rightarrow a+b+c\ge ab+bc+ca $ But by C-S we have: $ \frac{a}{2+a}+\frac{b}{2+b}+\frac{c}{2+c}=1\Leftrightarrow \frac{a^2}{2a+a^2}+\frac{b^2}{2b+b^2}+\frac{c^2}{2c+c^2}=1 $ $ \Leftrightarrow 1\ge\frac{(a+b+c)^2}{2(a+b+c)+a^2+b^2+c^2}\Leftrightarrow $ $ \Leftrightarrow a+b+c\ge ab+bc+ca\Leftrightarrow QED $ ____ Sandu Marin $ab+bc+ca+2abc=1\iff 1=\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$ $\Longrightarrow 1=\frac{a^2}{a^2+a}+\frac{b^2}{b^2+b}+\frac{c^2}{c^2+c}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2+a+b+c}$ $\iff a+b+c\ge 2(ab+bc+ca)=2-4abc \iff a+b+c+4abc\geq 2.$

Mister sqing, Our solutions are similar, if we consider the substitutions $ a\mapsto\frac{a}{2},b\mapsto\frac{b}{2},c\mapsto\frac{c}{2} $

Guys, me solution to the original problem. We have $ab + bc + ac + 2abc = 1$. Simply applying AM-GM gives us $abc \leq \frac{1}{8}$. Now, We have to prove that $2a + 2b + 2c + 1 \geq 32 abc$ Now as $2a + 2b + 2c + 1 \geq 4\sqrt[4]{8abc}$ its sufficient to prove that $4\sqrt[4]{8abc} \geq 32abc$ $\iff$ $\sqrt[4]{8abc} \geq 8abc$ $\iff$ $8abc \geq 8^4a^4b^4c^4$ $\iff$ $1 \geq 8^3 a^3 b^3 c^3 $ $\iff$ $1 \geq 8abc$ This is true by because of the given condition and hence we have proved the inequality. Moreover, equality occurs only when $2a=2b=2c=1$ or $a=b=c=\frac{1}{2}$ Please let me know if there is some error with the solution.

let $a=\frac{x}{y+z},b=\frac{y}{x+z},c=\frac{z}{x+y} $, then we $\sum ab+2abc=1\Leftrightarrow \sum \frac{xy}{y+z}(x+z)+2\frac{xyz}{\prod (x+y)}=1 $, it's true 1. $2(a+b+c)+1\geq 32abc\Leftrightarrow 2\sum \frac{x}{y+z}+1\geq 32\frac{xyz}{\prod (x+y)}\Leftrightarrow 2(\sum x(x+z)((x+y)))+\prod (x+y)\geq 32xyz $, denote by $p=x+y+z,q=xy+yz+zx,r=xyx $, then $2(p^{3}-2pq+3r)+pq-r\geq 32r\Leftrightarrow 2p^{3}-3pq-27r\geq 0\Leftrightarrow 2(p^{3}-4pq+9r)+5(pq-9r)\geq 0 $, schur's inequality

\[ 2\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)+1\ge 32\frac{xyz}{(y+z)(z+x)(x+y)}. \] sol. Denote by : $p=a+b+c,q=ab+bc+ca, r=abc $, then we $2(p^{3}-2pq+3r)+pq-r\geq 32r\Leftrightarrow 2(p^{3}-4pq+9r)+4pq+pq-9r\geq 0 $, schur's inequality

These solutions all look overkill. Here is a much simpler one. Let $m=1/a,n=1/b,k=1/c$. The problem is then equivalent to proving $2(mn+nk+mk)+mnk \geqslant 32$, under the condition $m+n+k+2=mnk$. Now, let $mnk\triangleq t^3$. Then by AM-GM, $t^3=m+n+k+2\geqslant 3t+2$, that is, $t\geqslant 2$, namely $mnk\geqslant 8$. Coming back to the original guy now, $mn+nk+mk\geqslant 3\sqrt[3]{m^2n^2k^2} = 3t^2 \geqslant 12$, and $mnk\geqslant 8$, yielding $2(mn+nk+mk)+mnk\geqslant 32$, as desired.