$\sum \frac{1}{a+b+1}\geq 1$ $\leftrightarrow$ $\sum (1-\frac{a+b}{a+b+1})\geq 1$ $\Leftrightarrow$ $2\geq \sum \frac{a+b}{a+b+1}=\sum \frac{(a+b)^{2}}{(a+b)^{2}+a+b}\geq \frac{4(a+b+c)^{2}}{2(\sum a^{2}+\sum ab+\sum a)}$ $\Rightarrow$ $\sum a^{2}+\sum ab+\sum a \geq (a+b+c)^{2}$ $\Leftrightarrow$ $a+b+c\geq ab+bc+ca$.qed

I have anther one. Let $a,b,c$ be positive real numbers such that $a+b+c=ab+bc+ca$. Prove that \[\frac1{1+a+b}+\frac1{1+b+c}+\frac1{1+c+a}\leq1.\]

?? That's in contradiction with the posted one...?

By applying the Cauchy-Schwarz inequality, we obtain \[ (a+b+1)(a+b+c^{2})\geq(a+b+c)^{2},\]thus we have \[\frac{1}{a+b+1}\leq\frac{c^{2}+a+b}{(a+b+c)^{2}}.\]Now by summing ciclically we obtain \[\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq\frac{a^{2}+b^{2}+c^{2}+2(a+b+c)}{(a+b+c)^{2}}\] But from the condition, we can see that \[ a^{2}+b^{2}+c^{2}+2(a+b+c)\geq (a+b+c)^{2},\]and therefore \[ a+b+c\geq ab+bc+ca.\] Equality occurs if and only if $a=b=c=1$.The ineq which pvthuan posted is also in my notes, and the solution is the same as above. Andrei Ciupan.

Otherwise: Let $a+b=x;b+c=y;c+a=z$. The inequality becomes: \[\sum\dfrac{1}{m+1}\ge 1\Rightarrow \sum (m+1)(n+1)\ge (m+1)(n+1)(p+1)\Rightarrow m+n+p+2\ge mnp\] We back at the variables $a,b,c$: $2(a+b+c)+2\ge (a+b)(b+c)(c+a)$. We have the identity : $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$. Note $a+b+c=m;ab+bc+ca=n$ and $abc=p$. By $AM-GM$, $mn\ge 9p$. But $2m+2\ge mn-p\ge \dfrac{8mn}{9}$. If $m<n$, we use the inequality : $m^2\ge 3n>3m\Rightarrow m>3\Rightarrow 9m+9\ge 4mn>12n>12m\Rightarrow m<3$, contradiction !

Romania Junior Balkan Team Selection Tests 2007 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=261698&p=2914767#p2914767 The following inequalities are also true. Let $a, b, c$ three positive reals such that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $ Show that\[a+b+c\geq 3. \]

sqing wrote: Romania Junior Balkan Team Selection Tests 2007 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=261698&p=2914767#p2914767 The following inequalities are also true. Let $a, b, c$ three positive reals such that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $ Show that \[a+b+c\geq 3. \] Prove :$1\ge \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\ge \frac{9}{2(a+b+c)+3}$

$abc=1\Rightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $ $a+b+c \geq ab+bc+ca \Rightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2924523 Positive real numbers $a,b,c$ satisfy$\frac{1}{2+a} + \frac{1}{2+b} + \frac{1}{2+c} = 1$Prove that$ a+b+c \geq ab+bc+ca$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2137595

Leta, b, c three positive reals such that 1/(a+b+1)+1/(c+b+1)+1/(a+c+1)≥ =>a+b+c≥a2b+b2a+c2b+b2c+c2a+a2c show that a+b+c≥ ab+bc+ac => a+b+c ≥a2b+b2a+a2c+c2a+b2c+c2b≥ab+bc+ac a2b+b2a≥2ab b2c+c2b≥2bc c2a+a2c≥2ac 2ab+2ac+2bc≥ab+bc+ca ab+bc+ac≥0

My solution: $ a,b,c>0,\sum\frac{1}{a+b+1}\ge 1\Leftrightarrow 2+2(a+b+c)+abc\ge (a+b+c)(ab+bc+ca) $ But, by Am-Gm, we have:$ abc\le\frac{(a+b+c)(ab+bc+ca)}{9} $; result $ 4(a+b+c)(ab+bc+ca)\le 9+9(a+b+c) $ It is sufficient that $ 9+9(a+b+c)\le 4(a+b+c)^2\Leftrightarrow a+b+c\ge 3 $ So for $ a+b+c\ge 3 $ inequality is true! If $ a+b+c\le 3 $ we have: $ 3(a+b+c)\ge (a+b+c)^2\ge 3(ab+bc+ca)\Rightarrow a+b+c\ge ab+bc+ca\Leftrightarrow QED $ The proof is ended! ________ Marin Sandu