let LHS=T now by Cauchy we have $2(a+b+c)T\geq (\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)})^2$ so we have to prove that : $\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}\geq 2$ Now use mixing to kill it!

K.N wrote: let LHS=T now by Cauchy we have $2(a+b+c)T\geq \frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}$ so we have to prove that : $\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}\geq 2$ Now use mixing to kill it! The $RHS$ of your first inequality should be $[\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}]^2$

K.N wrote: so we have to prove that $\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}\geq 2$ Let $a=\frac{x^2}{yz},b=\frac{y^2}{zx},c=\frac{z^2}{xy}$, then we need to prove that $\frac{\sum_{sym}{x^6y^3}-2\sum_{cyc}{x^5y^2z^2}}{\prod_{(cyc}{x^3+y^3+xyz)}}\geq 0$ which is obvious by Muirhead.

bgn wrote: K.N wrote: let LHS=T now by Cauchy we have $2(a+b+c)T\geq \frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}$ so we have to prove that : $\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}\geq 2$ Now use mixing to kill it! The $RHS$ of your first inequality should be $[\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}]^2$ Yes , sorry I changed it

Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ . Prove that: $$\frac {a}{(a+b+1)^2}+\frac {b}{(b+c+1)^2}+\frac {c}{(c+a+1)^2} \geq \frac {1}{a+b+c}$$$$\frac {a+b}{(a+b+3)^2}+\frac {b+c}{(b+c+3)^2}+\frac {c+a}{(c+a+3)^2} \geq \frac {18}{25(a+b+c)}$$

sqing wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ . Prove that: $$\frac {a}{(a+b+1)^2}+\frac {b}{(b+c+1)^2}+\frac {c}{(c+a+1)^2} \geq \frac {1}{a+b+c}$$ Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ . Prove that: $$\frac {a}{a+b+1}+\frac {b}{b+c+1}+\frac {c}{c+a+1} \geq \ 1 $$

Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ . Prove that: $$\frac {a}{a+b+1}+\frac {b}{b+c+1}+\frac {c}{c+a+1} \geq \ 1 $$Let $\left( {a;b;c} \right) = \left( {\frac{x}{y};\frac{y}{z};\frac{z}{x}} \right)$. $\sum {\frac{{\frac{x}{y}}}{{\frac{x}{y} + \frac{y}{z} + 1}}} = \sum {\frac{{xz}}{{xz + yz + {y^2}}}} = \sum {\frac{{{{\left( {xz} \right)}^2}}}{{{{\left( {xz} \right)}^2} + xy{z^2} + x{y^2}z}}} \ge \frac{{{{\left( {xy + yz + zx} \right)}^2}}}{{\sum {{x^2}{y^2}} + 2xyz\left( {x + y + z} \right)}} = 1$

What do you mean by mixing?

I mean mixing method replace $\sqrt {ab},\sqrt {ab} $ and you will get that the minimum occurres when $a=b=c$

bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ By Cauchy-Schwarz Inequality, we have \[\sum (a+b) \sum \frac{a+b}{(a+b+1)^2} \geqslant \left (\sum \frac{a+b}{a+b+1} \right )^2.\]Therefore, we need to show that \[\sum \frac{a+b}{a+b+1} \geqslant 2,\]equivalent to \[\sum \frac{1}{a+b+1} \leqslant 1.\]Which is USAMO 1997: http://artofproblemsolving.com/community/c6h948

Nguyenhuyen_AG wrote: bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ By Cauchy-Schwarz Inequality, we have \[\sum (a+b) \sum \frac{a+b}{(a+b+1)^2} \geqslant \left (\sum \frac{a+b}{a+b+1} \right )^2.\]Therefore, we need to show that \[\sum \frac{a+b}{a+b+1} \geqslant 2,\]equivalent to \[\sum \frac{1}{a+b+1} \leqslant 1.\]Which is USAMO 1997: http://artofproblemsolving.com/community/c6h948 Here is my solution for the last inequality Setting $a=x^3$ $b=y^3$ $c=z^3$ with $xyz$=1$ and here use x^3+y^3\ge xy(x+y)$ So İt remains to prove that $\frac{1}{xy(x+y)+1}+\frac{1}{yz(y+z)+1}+\frac{1}{zx(z+x)+1}\le 1$ but İt is equal to $1$.

Murad.Aghazade wrote: Nguyenhuyen_AG wrote: bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ By Cauchy-Schwarz Inequality, we have \[\sum (a+b) \sum \frac{a+b}{(a+b+1)^2} \geqslant \left (\sum \frac{a+b}{a+b+1} \right )^2.\]Therefore, we need to show that \[\sum \frac{a+b}{a+b+1} \geqslant 2,\]equivalent to \[\sum \frac{1}{a+b+1} \leqslant 1.\]Which is USAMO 1997: http://artofproblemsolving.com/community/c6h948 Here is my solution for the last inequality Setting $a=x^3$ $b=y^3$ $c=z^3$ with $xyz$=1$ and here use x^3+y^3\ge xy(x+y)$ So İt remains to prove that $\frac{1}{xy(x+y)+1}+\frac{1}{yz(y+z)+1}+\frac{1}{zx(z+x)+1}\le 1$ but İt is equal to $1$. Yes,but this solution is the one which was proposed on Nguyenhuyen's link. :/

Can you Share This link?

http://artofproblemsolving.com/community/c6h948

bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ sqing wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ . Prove that: $$\frac {a+b}{(a+b+3)^2}+\frac {b+c}{(b+c+3)^2}+\frac {c+a}{(c+a+3)^2} \geq \frac {18}{25(a+b+c)}$$ generalization Let $a,b,c,\lambda $ are positive real number such that $abc=1$ and $\lambda \ge 1$ . Prove that \[\frac{a+ b}{{{{\left( {a + b + \lambda } \right)}^2}}} + \frac{b + c}{{{{\left( {b + c + \lambda } \right)}^2}}} + \frac{c+ a}{{{{\left( {c + a + \lambda } \right)}^2}}} \ge \frac{{18}}{{{{\left( {\lambda + 2} \right)}^2}\left( {a + b + c} \right)}}{\rm{ }}\]

scpajmb wrote: bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ sqing wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ . Prove that: $$\frac {a+b}{(a+b+3)^2}+\frac {b+c}{(b+c+3)^2}+\frac {c+a}{(c+a+3)^2} \geq \frac {18}{25(a+b+c)}$$ generalization Let $a,b,c,\lambda $ are positive real number such that $abc=1$ and $\lambda \ge 1$ . Prove that \[\frac{a+ b}{{{{\left( {a + b + \lambda } \right)}^2}}} + \frac{b + c}{{{{\left( {b + c + \lambda } \right)}^2}}} + \frac{c+ a}{{{{\left( {c + a + \lambda } \right)}^2}}} \ge \frac{{18}}{{{{\left( {\lambda + 2} \right)}^2}\left( {a + b + c} \right)}}{\rm{ }}\] wow wow. What does $\lambda$ mean????

This conclusion is also true Let $a,b,c,\lambda $ are positive real number such that $\sqrt a + \sqrt b + \sqrt c = 3$ and $\lambda \ge 2$ . Prove that \[\frac{{a + b}}{{{{\left( {a + b + \lambda } \right)}^2}}} + \frac{{b + c}}{{{{\left( {b + c + \lambda } \right)}^2}}} + \frac{{c + a}}{{{{\left( {c + a + \lambda } \right)}^2}}} \ge \frac{{18}}{{{{\left( {\lambda + 2} \right)}^2}\left( {a + b + c} \right)}}{\rm{ }}\]

Nguyenhuyen_AG wrote: bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ By Cauchy-Schwarz Inequality, we have \[\sum (a+b) \sum \frac{a+b}{(a+b+1)^2} \geqslant \left (\sum \frac{a+b}{a+b+1} \right )^2.\]Therefore, we need to show that \[\sum \frac{a+b}{a+b+1} \geqslant 2,\]equivalent to \[\sum \frac{1}{a+b+1} \leqslant 1.\]Which is USAMO 1997: http://artofproblemsolving.com/community/c6h948 http://www.artofproblemsolving.com/community/c6h414996p2335872: $abc=1,a,b,c>0$,prove that , $$ \frac{1}{1+b+c}+ \frac{1}{1+a+c}+ \frac{1}{1+a+b}\le\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$$$(b)\ \ \ \sum_{cyc}\frac 1{a+b+1}\leq 1.$

bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ The following inequality is also true. Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: \[\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \leq \frac {2}{3}\]

oldbeginner wrote: bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ The following inequality is also true. Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: \[\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \leq \frac {2}{3}\] Very nice. Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that:$$\frac {2}{3}\geq\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$$

luofangxiang wrote: four short lines done!!! 四行秒杀型证明。

zdyzhj wrote: luofangxiang wrote: four short lines done!!! 四行秒杀型证明。 one short lines done!!!

One word done!!!

luofangxiang wrote: zdyzhj wrote: luofangxiang wrote: four short lines done!!! 四行秒杀型证明。 one short lines done!!! set $x=\frac{1}{a+b+1}, y=\frac{1}{b+c+1}, z=\frac{1}{c+a+1} $ with $x+y+z \le 1$ The inequality becomes $x^2(1-x)^2+y^2(1-y)^2+z^2(1-z)^2 \le \frac{27}{4}$ Note that by AM-GM $x(1-x)^2=4.x.\frac{(1-x)}{2}.\frac{(1-x)}{2} \le 4 (\frac{x+\frac{(1-x)}{2}+\frac{(1-x)}{2}}{3})^3=\frac{4}{27}$ Then $x^2(1-x)^2+y^2(1-y)^2+z^2(1-z)^2 \le \frac{4}{27} (x+y+z) \le \frac{4}{27}$

This is my train of though

luofangxiang wrote: zdyzhj wrote: luofangxiang wrote: four short lines done!!! 四行秒杀型证明。 one short lines done!!! we mean four short lines done is that we needn't any known result such as $\sum{\frac{1}{a+b+1}}\le 1$

scpajmb wrote: This conclusion is also true Let $a,b,c,\lambda $ are positive real number such that $\sqrt a + \sqrt b + \sqrt c = 3$ and $\lambda \ge 2$ . Prove that \[\frac{{a + b}}{{{{\left( {a + b + \lambda } \right)}^2}}} + \frac{{b + c}}{{{{\left( {b + c + \lambda } \right)}^2}}} + \frac{{c + a}}{{{{\left( {c + a + \lambda } \right)}^2}}} \ge \frac{{18}}{{{{\left( {\lambda + 2} \right)}^2}\left( {a + b + c} \right)}}{\rm{ }}\] After C-S, it becomes $\frac{a+b}{a+b+\lambda}+\frac{b+c}{b+c+\lambda}+\frac{c+a}{c+a+\lambda}\ge \frac{6}{\lambda+2}$ which is Vasc's old inequality.

scpajmb wrote: generalization Let $a,b,c,\lambda $ are positive real number such that $abc=1$ and $\lambda \ge 1$ . Prove that \[\frac{a+ b}{{{{\left( {a + b + \lambda } \right)}^2}}} + \frac{b + c}{{{{\left( {b + c + \lambda } \right)}^2}}} + \frac{c+ a}{{{{\left( {c + a + \lambda } \right)}^2}}} \ge \frac{{18}}{{{{\left( {\lambda + 2} \right)}^2}\left( {a + b + c} \right)}}{\rm{ }}\] After C-S, it becomes $\frac{a+b}{a+b+\lambda}+\frac{b+c}{b+c+\lambda}+\frac{c+a}{c+a+\lambda}\ge \frac{6}{\lambda+2}$ which is also Vasc's old inequality.

https://drive.google.com/file/d/0B1yzskQTMtL0dUFPQzV5bFdYRHc/view?usp=drivesdk

mehregan wrote: https://drive.google.com/file/d/0B1yzskQTMtL0dUFPQzV5bFdYRHc/view?usp=drivesdk Congratulations! Very nice and tricky solution

bgn wrote: Let $a,b,c \in \mathbb {R}^{+}$ and $abc=1$ prove that: $\frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2} \geq \frac {2}{a+b+c}$ $$\frac{a+b}{(a+b+1)^2}(a+b+c)=\frac{(a+b)^2}{(a+b+1)^2}+\frac{c(a+b)}{(a+b+1)^2}=\frac{(a+b)^2}{(a+b+1)^2}+\frac{a+b}{ab(a+b+1)^2}\geq$$$$\geq \frac{\left [\frac{1}{a+b+1}+\frac{a+b}{a+b+1}\right ]^2}{1+\frac{ab}{a+b}}=\frac{a+b}{a+b+ab}$$Now doing a cyclic shift $a\mapsto \frac{1}{a}$ changes nothing and hence we need to show that : $$\sum \frac{a+b}{a+b+1}\geq 2 \iff \sum \frac{1}{1+a+b}\geq 1$$$a:=\frac{yz}{x^2}$; $$\sum \frac{x^2y^2}{y^3z+x^3z+x^2y^2}\leq \sum \frac{x^2y^2}{zxy(x+y)+x^2y^2}=1$$And hence the desired.

Solution for the initial problem: assume wlog $a \ge b \ge c$ We will prove $\sum \frac{a+b}{(a+b+1)^2} \ge \sum \frac{a+b}{(a+b+c)^2}$ and then the result follows immediately. $\sum \frac{a+b}{(a+b+1)^2}-\frac{a+b}{(a+b+c)^2} $ $= \sum \frac{(a+b)(2a+2b+c+1)(c-1)}{((a+b+c)(a+b+1))^2}$ $\ge \frac{(b+c)(2b+2c+a)}{((a+b+c)(a+b+1))^2} \sum (c-1)$ (from $a \ge b \ge c$ we can obviously deduce to this) $=\frac{(b+c)(2b+2c+a)}{((a+b+c)(a+b+1))^2} (a+b+c-3)$ $\ge 0 $ (from $a+b+c \ge 3(abc)^{\frac{1}{3}} = 3$ since $abc=1$)

First, by Cauchy-Schwarz Inequality, we have \[(a+b+c)(a+b+ab)\geqslant (a+b+1)^2 \implies \frac{1}{(a+b+ab)^2}\geqslant \frac{1}{(a+b+c)(a+b+ab)}.\]It's then suffice to show that \[\sum_{cyc} \frac{a+b}{a+b+ab}\geqslant 2.\]By letting $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x},$ using Cauchy-Schwarz Inequality again, \[\sum_{cyc} \frac{a+b}{a+b+ab}=\sum_{cyc} \frac{xz}{xy+xz+y^2}+\sum_{cyc} \frac{y^2}{xy+xz+y^2}\geqslant \frac{(xz+yx+zy)^2}{\sum_{cyc} (x^2yz+x^2z^2+xy^2z)}+\frac{(x+y+z)^2}{\sum_{cyc} (xy+xz+y^2)}=2. \]

Let $T = \frac {a+b}{(a+b+1)^2}+\frac {b+c}{(b+c+1)^2}+\frac {c+a}{(c+a+1)^2}$. By applying cauchy on $T$ we have : $((a+b) + (b+c) + (c+a))(T) \ge (\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)})^2$ and $((a+b) + (b+c) + (c+a))\frac {2}{a+b+c} = 4$. so we want to prove $(\frac {a+b}{(a+b+1)}+\frac {b+c}{(b+c+1)}+\frac {c+a}{(c+a+1)}) \ge 2$ or $\frac{1}{a+b+1} + \frac{1}{b+c+1} + \frac{1}{c+a+1} \le 1$. $\frac{1}{a+b+1} + \frac{1}{b+c+1} + \frac{1}{c+a+1} = \frac{\sqrt[3]{abc}}{\sqrt[3]{a^3}+\sqrt[3]{b^3}+1} + \frac{\sqrt[3]{abc}}{\sqrt[3]{b^3}+\sqrt[3]{c^3}+1} + \frac{\sqrt[3]{abc}}{\sqrt[3]{c^3}+\sqrt[3]{a^3}+1} \le \frac{\sqrt[3]{abc}}{\sqrt[3]{a^2b}+\sqrt[3]{b^2a}+1} + \frac{\sqrt[3]{abc}}{\sqrt[3]{b^2c}+\sqrt[3]{c^2b}+1} + \frac{\sqrt[3]{abc}}{\sqrt[3]{c^2a}+\sqrt[3]{a^2a}+1} = \frac{\sqrt[3]{c}}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}} + \frac{\sqrt[3]{a}}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}} + \frac{\sqrt[3]{b}}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}} = 1$