I remember that I posted this inequality long long time ago. I have two solutions for this nice inequality. Anyway, here there are: Solution 1. The given inequality is equivalent with \[ \sum a(c+1)\geq\frac{3}{4} (1+a)(1+b)(1+c)\Leftrightarrow \] \[ \sum a+\sum ab\geq \frac{3}{4}(1+a)(1+b)(1+c)\Leftrightarrow \] \[ 4\sum a+4\sum ab\geq 3+3\sum a+3\sum ab+3abc\Leftrightarrow \] \[ a+b+c+ab+bc+ca\geq 6, \] a trivial application of AM-GM. Solution 2. We let $a=\frac{x}{y}, b=\frac{y}{z},c=\frac{z}{x}$. The inequality becomes \[ \sum\frac{\frac{x}{y}}{\left(\frac{x}{y}+1\right)\left(\frac{y}{z}+1\right)}\geq\frac{3}{4}\Leftrightarrow \] \[ \sum\frac{zx}{(x+y)(y+z)}\geq\frac{3}{4}. \] This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it ) or AM-GM.

Why integers?

Obviously it is a typo, and the problem is actually with positive reals. Cezar, we can also use the substitution $a = \frac 1x$, $b = \frac 1y$ and $c = \frac 1z$.

mathmanman wrote: Obviously it is a typo, and the problem is actually with positive reals. Yes indeed, I have edited the problem.

mathmanman wrote: Obviously it is a typo, and the problem is actually with positive reals. Cezar, we can also use the substitution $a = \frac 1x$, $b = \frac 1y$ and $c = \frac 1z$. I don't quite think so.

And, may I know why you "don't quite think so"? Actually, using this substitution gives that $\frac 1{xyz} = 1$. And then, if we put all on the same denominator, in the end, all we're left to prove is the exact same thing as in your first solution, that is : $yz + y + xz + z + xy + x \geq 6$, but, it can be rewritten (from the condition) : $x + \frac 1x + y + \frac 1y + z + \frac 1z \geq 6$, which is true. ($x + \frac 1x \geq 2$, a.s.o.)

But it is almost the same with my first solution.

Of course it is, but I just wanted to show you that we could use the substitution I said.. Oh, and, by the way, the equality case is : $a=b=c=1$.

supposing:$a\leq{b}\leq{c}$ According to the inequality of réordonnement we have: $\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)}=W$ and:$\frac{1}{(b+1)(c+1)}\leq{\frac{1}{(c+1)(a+1)}}\leq{\frac{1}{(a+1)(b+1)}}$ According to the inequality of shebyshev: $W\geq{\frac{1}{3}\((a+b+c)\((\frac{3+a+b+c}{(1+a)(1+b)(1+c)})}$ supposing : $\frac{a+b+c}{3}=\alpha$ according to the inequalities between averages : $3\alpha=a+b+c\geq3{(abc)}^{1/3}=3$ and, $(1+a)(1+b)(1+c)\leq{[\frac{(1+a)+(1+b)+(1+c)}{3}]}^{3}={(1+\alpha)}^{3}$ so: $W\geq\alpha\frac{6}{{(1+\alpha)}^{3}}$ it is enough to prove that: $\frac{6\alpha}{{(1+\alpha)}^{3}}\geq\frac{3}{4}$ we know that $\alpha\geq{1}$and$f(\alpha)=\frac{6\alpha}{{(1+\alpha)}^{3}}$ is strictly increasing on $R^{+}$ we have also $f(\alpha)\geq{f(1)}=\frac{3}{4}\longleftrightarrow{W}\geq{\frac{3}{4}}$

let $a=m/n, b=n/p, c=p/m$... then, the inequality is equivalent to $\sum_{cyc}\frac{mp}{(m+n)(p+n)}\geq \frac{3}{4}$ $\Leftrightarrow 4\sum_{cyc}mp(m+p)\geq (3\sum_{cyc}m^{2}p+mp^{2})+6mnp$ $\Leftrightarrow \sum_{cyc}mp(m+p)\geq 6mnp$ which is evidently true... equality holds iff $m=n=p$, i.e iff $a=b=c=1$. edit: i'm sorry... i didn't see that cezar lupu had proposed the same solution...

An interesting generalization is the following: Find the largest positive real number $p$ such that \[ \frac{a}{(a+1)(b+p)}+\frac{b}{(b+1)(c+p)}+\frac{c}{(c+1)(a+p)}\geq \frac{3}{2(1+p)}, \] for any positive real numbers $a,b,c$ such that $abc=1$.

I think i have a straightout Muirhead solution that is not bad at all. We start with $ \dfrac{a}{(a+1)(b+1)}+\dfrac{b}{(b+1)(c+1)}+ \dfrac{c}{(c+1)(a+1)} \geq \dfrac{3}{4}$. Multiply the common denominator $ 4(a+1)(b+1)(c+1)$. This yields $ 4(ab+bc+ca) + 4(a+b+c) \ge 3(a+b+c) + 3(ab+bc+ca) + 3abc + 3$. This becomes $ \frac{1}{2}T[1,1,0] + \frac{1}{2}T[1,0,0] \ge \frac{1}{2}T[1,1,1]+\frac{1}{2}T[0,0,0]$. Since $ abc = 1$, we have this equivalent to $ \frac{1}{2}T[1,1,0] + \frac{1}{2}T[1,0,0] \ge \frac{1}{2}T[\frac{2}{3},\frac{2}{3},\frac{2}{3}]+\frac{1}{2}T[\frac{1}{3},\frac{1}{3},\frac{1}{3}]$ and the result follows by Muirhead. The equality occurs if and only if $ a = b = c = 1$.

For: $ a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z},$ we get to: $ LHS=\frac{y^2}{(x+y)(y+z)}+\frac{z^2}{(y+z)(z+x)}+\frac{x^2}{(z+x)(x+y)}\stackrel{Titu's\ lemma}{\geq}$ $ \geq\frac{(x+y+z)^2}{(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)}=$ $ =\frac{x^2+y^2+z^2+2(xy+yz+zx)}{x^2+y^2+z^2+3(xy+yz+zx)}\stackrel{(*)}{\geq}\frac{3}{4}=RHS$ We took into account that $ (*)$ is equivalent to: $ 4(x^2+y^2+z^2)+8(xy+yz+zx)\geq3(x^2+y^2+z^2)+9(xy+yz+zx)\ \Leftrightarrow$ $ \ \Leftrightarrow\ x^2+y^2+z^2\geq xy+yz+zx.$ Obviously, we have equality for: $ x=y=z\ \Leftarrow\ \ a=b=c=1$ For Vasc's inequality I found that $ p=2$. Is it right, Vasile? Marin Bancoș - ROMANIA

marin.bancos wrote: For Vasc's inequality I found that $ p = 2$. Is it right, Vasile? Marin Bancoș - ROMANIA Yes, it is right.

Vasc wrote: An interesting generalization is the following: Find the largest positive real number $ p$ such that \[ \frac {a}{(a + 1)(b + p)} + \frac {b}{(b + 1)(c + p)} + \frac {c}{(c + 1)(a + p)}\geq \frac {3}{2(1 + p)},\] for any positive real numbers $ a,b,c$ such that $ abc = 1$. See my proof here, Mr Vasile: http://canhang2007.wordpress.com/2009/11/08/inequality-54-v-cirtoaje/

Similarly, we can show that $ \sum \frac a{(a+p)(b+q)} \ge \frac 3{(1+p)(1+q)}$ for any $ q\ge \frac 1{2}$ and $ p\ge \frac {2q-1}{q+1}$.

Cezar Lupu wrote: I remember that I posted this inequality long long time ago. I have two solutions for this nice inequality. Anyway, here there are: Solution 1. The given inequality is equivalent with \[ \sum a(c + 1)\geq\frac {3}{4} (1 + a)(1 + b)(1 + c)\Leftrightarrow\] \[ \sum a + \sum ab\geq \frac {3}{4}(1 + a)(1 + b)(1 + c)\Leftrightarrow\] \[ 4\sum a + 4\sum ab\geq 3 + 3\sum a + 3\sum ab + 3abc\Leftrightarrow\] \[ a + b + c + ab + bc + ca\geq 6,\] a trivial application of AM-GM. Solution 2. We let $ a = \frac {x}{y}, b = \frac {y}{z},c = \frac {z}{x}$. The inequality becomes \[ \sum\frac {\frac {x}{y}}{\left(\frac {x}{y} + 1\right)\left(\frac {y}{z} + 1\right)}\geq\frac {3}{4}\Leftrightarrow\] \[ \sum\frac {zx}{(x + y)(y + z)}\geq\frac {3}{4}.\] This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it ) or AM-GM. how to solve it using am-gm?

sorry, that was nonsense

Martin N. wrote: tkrass wrote: Cezar Lupu wrote: \[ \sum\frac {zx}{(x + y)(y + z)}\geq\frac {3}{4}.\] This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it ) or AM-GM. how to solve it using am-gm? \[ \sum\frac {zx}{(x + y)(y + z)}\geqslant\frac {3}{4}\Longleftrightarrow \sum\frac {zx}{\frac {x + y}{2}\cdot\frac {y + z}{2}}\geqslant 3\] With $ \frac {x + y}{2}\geqslant\sqrt {xy}$, we have to prove \[{ \sum\frac {zx}{y\sqrt {zx}} = \sum\frac {\sqrt {zx}}{y}}\geqslant 3\mbox{,}\] but this is obvious as, by AM-GM again, \[ \frac {\sum\frac {\sqrt {zx}}{y}}{3}\geqslant\sqrt [3]{\prod\frac {\sqrt {zx}}{y}} = 1\] Martin N. your proof is wrong.

Vasc wrote: An interesting generalization is the following: Find the largest positive real number $p$ such that \[ \frac{a}{(a+1)(b+p)}+\frac{b}{(b+1)(c+p)}+\frac{c}{(c+1)(a+p)}\geq \frac{3}{2(1+p)}, \] for any positive real numbers $a,b,c$ such that $abc=1$. How could we find the solutions or your generalisations? (both aren't solved yet with proof)

$\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)}\ge3/4$ I found something interesting. If we assume the statement to be true then I find that $a(c+1)+b(a+1)+c(b+1)\ge3/4(abc+ab+ac+bc+a+b+c+1)$ $ab+ac+bc+a+b+c\ge3abc+3$ $a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\ge6$ WE know by AM-GM $x+\frac{1}{x}\ge2$ NOW we are done.

Take $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$. Inequality becomes: $\sum_{cyc}\frac{xz}{(x+y)(y+z)}=\frac{\sum_{cyc}xz(x+z)}{(x+y)(y+z)(z+x)}\geq\frac{3}{4}$. $4(\sum_{symmetric}xy^2)\geq 3(2xyz+\sum_{symmetric}xy^2)$ $\sum_{symmetric}xy^2\geq 6xyz$, witch is true because of AM-GM.

sorry.....

$a+1\geq 2\sqrt{a}$ etc, we have $(a+1)(b+1)(c+1)\geq 8abc$, yielding $(a+1)(b+1)(c+1)\geq 8.$ $\therefore \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}$ $=1-\frac{2}{(a+1)(b+1)(c+1)}\geq 1-2\cdot\frac 18=\frac 34.$

kunny wrote: $a+1\geq 2\sqrt{a}$ etc, we have $(a+1)(b+1)(c+1)\geq 8abc$, yielding $(a+1)(b+1)(c+1)\geq 8.$ $\therefore \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}$ $=1-\frac{2}{(a+1)(b+1)(c+1)}\geq 1-2\cdot\frac 18=\frac 34.$ beautiful. Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove that \[\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}. \] Have a similar simple proof?

sqing wrote: kunny wrote: $a+1\geq 2\sqrt{a}$ etc, we have $(a+1)(b+1)(c+1)\geq 8abc$, yielding $(a+1)(b+1)(c+1)\geq 8.$ $\therefore \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}$ $=1-\frac{2}{(a+1)(b+1)(c+1)}\geq 1-2\cdot\frac 18=\frac 34.$ beautiful. Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove that \[\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}. \] Have a similar simple proof? Thanks^^ $\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)}$ $\geq \frac{ab+bc+ca+abc}{(a+1)(b+1)(c+1)}$ $=1-\frac{2}{(a+1)(b+1)(c+1)}\geq \frac 34.$

Thank kunny. $\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)}$ $=\frac{a^2+b^2+c^2+a+b+c}{(a+1)(b+1)(c+1)}$ $\geq \frac{ab+bc+ca+a+b+c+abc-1}{(a+1)(b+1)(c+1)}$ $=1-\frac{2}{(a+1)(b+1)(c+1)}\geq \frac 34.$

Cezar Lupu wrote: I remember that I posted this inequality long long time ago. I have two solutions for this nice inequality. Anyway, here there are: Solution 1. The given inequality is equivalent with \[ \sum a(c+1)\geq\frac{3}{4} (1+a)(1+b)(1+c)\Leftrightarrow \] \[ \sum a+\sum ab\geq \frac{3}{4}(1+a)(1+b)(1+c)\Leftrightarrow \] \[ 4\sum a+4\sum ab\geq 3+3\sum a+3\sum ab+3abc\Leftrightarrow \] \[ a+b+c+ab+bc+ca\geq 6, \] a trivial application of AM-GM. Solution 2. We let $a=\frac{x}{y}, b=\frac{y}{z},c=\frac{z}{x}$. The inequality becomes \[ \sum\frac{\frac{x}{y}}{\left(\frac{x}{y}+1\right)\left(\frac{y}{z}+1\right)}\geq\frac{3}{4}\Leftrightarrow \] \[ \sum\frac{zx}{(x+y)(y+z)}\geq\frac{3}{4}. \] This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it ) or AM-GM. Your first solution is very beautiful and very natural. Congratulations!

sqing wrote: Thank kunny. $\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)}$ $=\frac{a^2+b^2+c^2+a+b+c}{(a+1)(b+1)(c+1)}$ $\geq \frac{ab+bc+ca+a+b+c+abc-1}{(a+1)(b+1)(c+1)}$ $=1-\frac{2}{(a+1)(b+1)(c+1)}\geq \frac 34.$ You are welcome. That's exactly same as my solution.

Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$.Then, Clearing the denominator, we get $6S(4,2,0)+6S(4,1,1)+6S(3,3,0) \geq 12S(3,2,1)+6S(2,2,2)$ Obviously this holds by Muirhead inequality

Igor wrote: Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \] When is there equality? $$\iff$$$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$Where $a,b,c$ be three positive real numbers such that $abc=1$. 2005 Czech here

sqing wrote: Igor wrote: Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \] When is there equality? $$\iff$$$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$Where $a,b,c$ be three positive real numbers such that $abc=1$. It's obvious that the problems are not equivalent. Greece implies Czech,but the other implication is not true.

mihaig wrote: sqing wrote: Igor wrote: Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \] When is there equality? $$\iff$$$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$Where $a,b,c$ be three positive real numbers such that $abc=1$. It's obvious that the problems are not equivalent. Greece implies Czech,but the other implication is not true. Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \]$$\iff$$$$\frac{a}{(a+1)(c+1)}+\frac{c}{(c+1)(b+1)}+ \frac{b}{(b+1)(a+1)} \geq \frac{3}{4}$$$$\iff$$$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$Where $a,b,c$ be three positive real numbers such that $abc=1$.

Really? Are you sure that is the meaning of "equivalent"? Didn't realize that abc=1 changes all in my inequality? Because it's non-homogenuous,so it's not allowed to replace by 1 how you wish and to claim the equivalency. For example,why didn't you replace all 1's from Czech by abc,then in my inequality again. Rookie mistakes and affirmations.

Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \]$$\iff$$$$\frac{y^2}{(x+y)(y+z)}+\frac{z^2}{(y+z)(z+x)}+\frac{x^2}{(z+x)(x+y)}{\geq} \frac{3}{4}. $$Where $x,y,z>0.$ Moscow

Indeed mihaig, you're right as the second inequality should be replaced with $\frac{xz}{(x+y)(y+z)}+\frac{xy}{(y+z)(z+x)}+\frac{yz}{(z+x)(x+y)}{\geq} \frac{3}{4}. $

WolfusA wrote: Indeed mihaig, you're right as the second inequality should be replaced with $\frac{xz}{(x+y)(y+z)}+\frac{xy}{(y+z)(z+x)}+\frac{yz}{(z+x)(x+y)}{\geq} \frac{3}{4}. $ Thank you. I wish I wasn't right...

Let $a=x/y,b=y/z,$ and $c=z/x.$ Since $(2,1,0)\succ(1,1,1),$ Muirhead yields \begin{align*}\sum_{\text{sym}}x^2y\ge6xyz&\implies4\sum_{\text{sym}}x^2y\ge3\left(2xyz+\sum_{\text{sym}}x^2y\right)\\&\implies 4\sum_{\text{cyc}}xz(x+z)\ge3(x+y)(y+z)(z+x)\\&\implies\sum_{\text{cyc}}\frac{xz(x+z)}{(x+y)(y+z)(z+x)}\ge\frac{3}{4}.\end{align*}Noting $$\frac{a}{(a+1)(b+1)}=\frac{\frac{x}{y}}{\frac{x+y}{y}\cdot\frac{y+z}{z}}=\frac{xz}{(x+y)(y+z)}$$finishes. $\square$