sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ Guess who is the author of the problem? It's taken from the BMOJ Shortlist 2018,which held in Greece. Thanks for posting!

I'm sorry what exactly is the pre-selection? And is the 2018 Shortlist available anywhere?

Steve12345 wrote: I'm sorry what exactly is the pre-selection? And is the 2018 Shortlist available anywhere? Of course it's not. But as in 2018,when Azerbaidjan used another problem of mine from the 2017 SHL,any country from the circuit has the right to use the problems from the previous year. Thus,due to sqing and the Elen Olympic Comitee,I found today that my problem is in 2018 SHL.

That a really nice problem.

Thank you,Hamel!

I'm still a bit confused. This is mihaig's problem, and was selected in the 2018 JBMO shortlist?

Steve12345 wrote: I'm still a bit confused. This is mihaig's problem, and was selected in the 2018 JBMO shortlist? It's one of the Romania proposasl for Juniors in 2018 with author mihaig,yes. I suppose it's in the SHL for Juniors,otherwise I can't explain the mistery. I never posted it,so it's taken from a list of problems.

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ $$\iff$$$$(a+b+c+ab+bc+ca)(1-abc+a^2b^2c^2)\geq3abc(1+abc).$$$$(a+b+c+ab+bc+ca)(1-abc+a^2b^2c^2)\geq 6\sqrt{abc}(1-abc+a^2b^2c^2)\geq \frac{12abc(1-abc+a^2b^2c^2)}{1+abc}\geq3abc(1+abc).$$(socrates)

Of course,the official solution (mine) is the same. But it takes few steps to get there.Try solve it by yourself.

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$

Bravo! It's elegant.

luofangxiang wrote: sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ $$(a+b+c+ab+bc+ca)(1-abc+a^2b^2c^2)\geq 6\sqrt{abc}(1-abc+a^2b^2c^2)\geq \frac{12abc(1-abc+a^2b^2c^2)}{1+abc}\geq3abc(1+abc).$$$$\frac{12abc(1-abc+a^2b^2c^2)}{1+abc}\geq3abc(1+abc)\iff (abc-1)^2\geq 0.$$(socrates)

Hea! It's the official solution,Giugiuc.

Solution without expanding: Let $abc=k^3$ and set $a=k\dfrac{x}{y}$, $b=k\dfrac{y}{z}$, $c=k\dfrac{z}{x}$, where $k,x,y,z>0$. Then, the inequality can be rewritten as $$\frac{z^2}{(ky+z)(kz+x)}+\frac{x^2}{(kz+x)(kx+y)}+\frac{y^2}{(kx+y)(ky+z)}\geq\frac{3k^2}{(1+k^3)^2}.$$Using the Cauchy-Schwarz inequality we have that $$\sum_{cyclic}\frac{z^2}{(ky+z)(kz+x)}\geq \frac{(x+y+z)^2}{(ky+z)(kz+x)+(kz+x)(kx+y)+(kx+y)(ky+z)},$$therefore it suffices to prove that $$\frac{(x+y+z)^2}{(ky+z)(kz+x)+(kz+x)(kx+y)+(kx+y)(ky+z)}\geq \frac{3k^2}{(1+k^3)^2}$$or $$\left((1+k^3)^2-3k^3\right)(x^2+y^2+z^2)\geq \left(3k^2(k^2+k+1)-2(1+k^3)^2\right)(xy+yz+zx).$$Since $x^2+y^2+z^2\geq xy+yz+zx$ and $(1+k^3)^2-3k^3>0$, it is enough to prove that $$(1+k^3)^2-3k^3\geq 3k^2(k^2+k+1)-2(1+k^3)^2,$$or $$(k-1)^2(k^2+1)(k+1)^2\geq 0,$$which is true.

Let $a,b,c$ be positive real numbers . Prove that $$ \frac{a}{b^2(b+1)(c+1)}+\frac{b}{c^2(c+1)(a+1)}+\frac{c}{a^2(a+1)(b+1)}\geq\frac{a+b+c}{(1+abc)^2}.$$$$ \frac{1}{a^2(a+1)(b+1)}+\frac{1}{b^2(b+1)(c+1)}+\frac{1}{c^2(c+1)(a+1)}\geq\frac{3}{(1+abc)^2}.$$Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \]2005 Czech 2006 France https://artofproblemsolving.com/community/c6h1439313p8176400

silouan wrote: Solution without expanding: Let $abc=k^3$ and set $a=k\dfrac{x}{y}$, $b=k\dfrac{y}{z}$, $c=k\dfrac{z}{x}$, where $k,x,y,z>0$. Then, the inequality can be rewritten as $$\frac{z^2}{(ky+z)(kz+x)}+\frac{x^2}{(kz+x)(kx+y)}+\frac{y^2}{(kx+y)(ky+z)}\geq\frac{3k^2}{(1+k^3)^2}.$$Using the Cauchy-Schwarz inequality we have that $$\sum_{cyclic}\frac{z^2}{(ky+z)(kz+x)}\geq \frac{(x+y+z)^2}{(ky+z)(kz+x)+(kz+x)(kx+y)+(kx+y)(ky+z)},$$therefore it suffices to prove that $$\frac{(x+y+z)^2}{(ky+z)(kz+x)+(kz+x)(kx+y)+(kx+y)(ky+z)}\geq \frac{3k^2}{(1+k^3)^2}$$or $$\left((1+k^3)^2-3k^3\right)(x^2+y^2+z^2)\geq \left(3k^2(k^2+k+1)-2(1+k^3)^2\right)(xy+yz+zx).$$Since $x^2+y^2+z^2\geq xy+yz+zx$ and $(1+k^3)^2-3k^3>0$, it is enough to prove that $$(1+k^3)^2-3k^3\geq 3k^2(k^2+k+1)-2(1+k^3)^2,$$or $$(k-1)^2(k^2+1)(k+1)^2\geq 0,$$which is true. Beautiful! P.S. Thanks for choosing my problem to be there. Posted here too. https://www.facebook.com/photo.php?fbid=1436710413133056&set=a.427827570688017&type=3&theater

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ Notice that the inequality proposed by me implies Czech,while Czech 2005 doesn't imply my inequality.Think! Now the form in which I sent the problem.

$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$is like proving $$(a+1)(b+1)(c+1)\geq\frac{(1+abc)^4}{1+(abc)^3}.$$ Using Holder inequality, we get: $$(a+1)(b+1)(c+1)((abc)^3+1)\geq (abc+1)^4.$$The problem is done. Thanks for the correction.

Interesting approach. But there is something wrong. Check again.

mihaig wrote: Guess who is the author of the problem? It's taken from the BMOJ Shortlist 2018,which held in Greece. Thanks for posting! Some persons make regrettable (and mean as well) confusions. We'll show that Czech 2005 doesn't imply Greece 2019. Consider the inequality $ \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 $a,b,c>0$ and $abc=1$. According to a poster, we can only make $\frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq \frac{3}{(1+abc)^2}$. But what law of mathematics forbids us to make ,say, $\frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4(abc)^2}$,which of course is generally false because it's equivalent to $\frac{a+b+c+ab+bc+ca}{(a+1)(b+1)(c+1)}\geq \frac{3}{4abc}$, where we may set $b=c=1,a \to 0$ to conclude that it is false.

sqing wrote: sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ $$\iff$$$$(a+b+c+ab+bc+ca)(1-abc+a^2b^2c^2)\geq3abc(1+abc).$$$$(a+b+c+ab+bc+ca)(1-abc+a^2b^2c^2)\geq 6\sqrt{abc}(1-abc+a^2b^2c^2)\geq \frac{12abc(1-abc+a^2b^2c^2)}{1+abc}\geq3abc(1+abc).$$(socrates) musan1909 wrote: $$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$is like proving $$(a+1)(b+1)(c+1)\geq\frac{(1+abc)^4}{1+(abc)^3}.$$ Using Holder inequality, we get: $$(a+1)(b+1)(c+1)((abc)^3+1)\geq (abc+1)^4.$$The problem is done. Thanks for the correction. Very nice solutions. $$(a+b+c+ab+bc+ca)(1-abc+a^2b^2c^2)\geq3abc(1+abc) \iff (a+1)(b+1)(c+1)((abc)^3+1)\geq (abc+1)^4 $$

https://artofproblemsolving.com/community/c6h6044p820582 Here's an old and stronger one.

Let $a,b,c$ be positive real numbers . Then$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{\sqrt[3]{(abc)^2}(1+\sqrt[3]{abc})^2}\geq\frac{3}{(1+abc)^2}.$$

ywq233 wrote: https://artofproblemsolving.com/community/c6h6044p820582 Here's an old and stronger one. Stronger than what? To what post from the link are you referring ?

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ Let $abc=w^3$. Thus, we need to prove that$$\frac{\sum\limits_{cyc}(a+ab)}{\prod\limits_{cyc}(a+1)}\geq\frac{3abc}{(1+abc)^2}$$or $$\frac{\sum\limits_{cyc}(a+ab)}{\prod\limits_{cyc}(a+1)}-1\geq\frac{3abc}{(1+abc)^2}-1$$or $$-\frac{1+abc}{\prod\limits_{cyc}(1+a)}\geq-\frac{1-abc+a^2b^2c^2}{(1+abc)^2}$$and since by Holder $$\prod\limits_{cyc}(1+a)\geq(1+\sqrt[3]{abc})^3=(1+w)^3,$$it's enough to prove that $$-\frac{1+w^3}{(1+w)^3}\geq-\frac{1-w^3+w^6}{(1+w^3)^2}$$or $$\frac{1-w^3+w^6}{(1+w^3)^2}\geq\frac{1-w+w^2}{(1+w)^2}$$or $$w(1-w^2)^2(1+w^2)\geq0.$$

sqing wrote:

Let $a,b,c$ be positive real numbers . Then$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{\sqrt[3]{(abc)^2}(1+\sqrt[3]{abc})^2}\geq\frac{3}{(1+abc)^2}.$$ What a beauty! I can't wait to see the author's solution.

arqady wrote: sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ Let $abc=w^3$. Thus, we need to prove that$$\frac{\sum\limits_{cyc}(a+ab)}{\prod\limits_{cyc}(a+1)}\geq\frac{3abc}{(1+abc)^2}$$or $$\frac{\sum\limits_{cyc}(a+ab)}{\prod\limits_{cyc}(a+1)}-1\geq\frac{3abc}{(1+abc)^2}-1$$or $$-\frac{1+abc}{\prod\limits_{cyc}(1+a)}\geq-\frac{1-abc+a^2b^2c^2}{(1+abc)^2}$$and since by Holder $$\prod\limits_{cyc}(1+a)\geq(1+\sqrt[3]{abc})^3=(1+w)^3,$$it's enough to prove that $$-\frac{1+w^3}{(1+w)^3}\geq-\frac{1-w^3+w^6}{(1+w^3)^2}$$or $$\frac{1-w^3+w^6}{(1+w^3)^2}\geq\frac{1-w+w^2}{(1+w)^2}$$or $$w(1-w^2)^2(1+w^2)\geq0.$$ Thank you! Respect

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ I think this is a different sol. The inequality is homogenous, so let $abc = 1$. Multiply both sides by $(a+1)(b+1)(c+1)$ \begin{align} \sum_{cyc}\frac{a+1}{ab} &\geq \frac{3(a+1)(b+1)(c+1)}{(1+abc)^2} \\ \sum_{cyc} ac+c &\geq \frac{3}{4} (abc + ab + bc + ac + a + b + c + 1) \\ 4(ab + ac + bc + a + b + c) & \geq 3(abc + ab + bc + ac + a + b + c + 1) \\ (ab + bc + ac + a + b + c) &\geq 3abc + 3 \\ \end{align} The end is true by AM-GM

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ sqing wrote: Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \]2005 Czech 2006 France So, lets start performing by an algebraic transformation i.e. $a\to \frac{x}{y}, b\to \frac{y}{z}, c \to \frac{z}{x}.$ Thus the inequality is equivalent to $$ \frac{1}{\frac{x}{y}\cdot \frac{y}{z}(\frac{y}{z}+1)(\frac{z}{x}+1)}+\frac{1}{\frac{y}{z}\cdot \frac{z}{x}(\frac{z}{x}+1)(\frac{x}{y}+1)}+\frac{1}{\frac{x}{y}\cdot \frac{z}{x}(\frac{x}{y}+1)(\frac{y}{z}+1)}\geq\frac{3}{(1+\frac{x}{y}\cdot \frac{y}{z}\cdot \frac{z}{x})^2}.$$or $$ \frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq\frac{3}{4}.$$or $$4\cdot\sum_{sym}x^2y^1z^0\geqslant 3\cdot \prod_{cyc}(x+y).$$or $$\sum_{sym}x^2y^1z^0\geqslant 6xyz$$which is obvious by AM-GM Inequality. $\blacksquare$ Thank you, best regards !

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ See my solution here.

We need to show the equivalent inequality \[(1+abc)^2\sum_{cyc}a(b+1)\geq 3abc(a+1)(b+1)(c+1)\]which is \[(a^2b^2c^2-abc+1)\sum_{cyc}a(b+1)\geq 3abc(abc+1).\]Note since $a^2b^2c^2+1>abc$, let $abc=t$, then use AM-GM, \[\sum_{cyc}a(b+1)\geq 6\sqrt{t}\]and so it is just showing \[6(t^2-t+1)\sqrt{t}\geq 3t(t+1)\]which follows from \[(t-1)^2(4t^2-t+4)\geq 0.\]And equality occurs at $a=b=c=1. \square$

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ I think the problem created from \[\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \]Indeed, using known inequality $(x+y+z)^2 \geqslant 3(xy+yz+zx),$ we have \[\left(\sum \frac{1}{a\left(b+1\right)}\right)^2 \geqslant 3\sum\frac{1}{ab(b+1)(c+1)} \geqslant \frac{9}{(abc+1)^2}.\]See here: https://artofproblemsolving.com/community/c6h6044p820582

sqing wrote: Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$ Let $x=\sum_{cyc} ab+\sum_{cyc} a$ and $y=abc$. Note that $$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}=\frac{x}{y(a+1)(b+1)(c+1)}=\frac{x}{y(x+y+1)}$$Thus we only need to prove that $$\frac{x}{y(x+y+1)}\ge \frac{3}{(1+y)^2} \iff x\ge \frac{3y^2+3y}{y^2-y+1}$$Let $f(a,b,c)=x=\sum_{cyc} ab+\sum_{cyc} a$ and $y=g(a,b,c)=abc$. Note that $$f(a,b,c)-f(\sqrt{ab},\sqrt{ab},c)=(c+1)(\sqrt{a}-\sqrt{b})^2 \ge 0 \implies f(a,b,c)\ge f(\sqrt{ab},\sqrt{ab},c)$$and $g(a,b,c)=g(\sqrt{ab},\sqrt{ab},c)$. Hence by SMV, we only need to prove $x\ge \frac{3y^2+3y}{y^2-y+1}$ for $a=b=c$ for which, using $x=3a^2+3a,y=a^3$, it is equivalent to $$3 (a + 1)^3 a (a - 1)^2 (a^2 + 1) \ge 0$$which is obviously true. $\blacksquare$

Bravo, nice proof.

mihaig wrote: Bravo, nice proof. ideed nice solution,beautiful problem!

Thank you very much

Let $p=a+b+c$ ; $q=ab+bc+ca$ ;$r=abc$ $(q+p)(1+r)^2 \geq 3r(p+q+r+1)$ $(q+p)(1+r^2) \geq (3r^2+rp+qr+3r)$ $(q+p)(1+r)^2 \geq 6r^2+2rp+2rq+6r$ $(q+p)(1+r^2) \geq 6r^2+6r$ $(q+p)(1+r) \geq 12r$ $(p+\frac{9r}{p})(2\sqrt{r}) \geq 4(\sqrt{9r})(\sqrt{r}) =12r$

Jay17 wrote: The inequality is homogenous, so let $abc = 1$. The inequality is not homogenous.

I'm not lookijg at this problem greece way too gay.