really hard: strong bound for cyclic sum 1 / (a(b+1)) + ...

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really hard: strong bound for cyclic sum 1 / (a(b+1)) + ...

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Source: my teacher In Taiwan's IMO training camp; "Old and New Inequalities"

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38 posts

#1 h Jun 8, 2004, 8:14 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Let $a$ , $b$ , $c$ be positive real numbers.

Prove that:

$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{\sqrt[3]{abc}\cdot\left(\sqrt[3]{abc}+1\right)}$ .

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5526 posts

harazi

#2 h Jun 8, 2004, 8:36 AM • 5 Y

Y by Adventure10, Mango247, and 3 other users

I mean, this is not extremely hard:
Since
$\frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}$
$=(1+abc)(\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)})+3$ ,
we can write the desired inequality in the form $\frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}\geq \frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}$ . Now, break $\frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}$ into
$\frac{1+a}{ab+a}+\frac{b+1}{bc+b}+\frac{c+1}{ca+c}\geq\frac{3}{\sqrt[3]{abc}}$ and $\frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1}\geq 3\sqrt[3]{abc}$ , and it is proven.

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#3 h Jun 8, 2004, 4:24 PM • 2 Y

Y by Adventure10, Mango247

Oh my..
I understand now,thanks a lot!!

But I dont think its "easy" to notice
that we can write another form like this...

BTW, this solution is so nice

I will never think about this~~
thanks your help

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6555 posts

#4 h Dec 13, 2005, 5:18 PM • 2 Y

Y by Adventure10 and 1 other user

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 for a weaker inequality:

$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{abc+1}$ .

Note that some proofs of this weaker inequality can be easily modified to prove the above inequality

$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{\sqrt[3]{abc}\left(\sqrt[3]{abc}+1\right)}$ .

Darij

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#5 h Aug 28, 2006, 3:19 AM • 2 Y

Y by Adventure10, Mango247

The keypoint of my solution is to set $a=t \frac{q}{p}, b=t\frac{r}{q}, c=t\frac{p}{r}$ where $t^{3}=abc$ .

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3952 posts

#6 h Aug 28, 2006, 12:35 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The solution I know use only AM-GM .
We have $(1+abc)(\sum\frac{1}{a(1+b)})+3=\sum\frac{1+a}{a(1+b)}+\sum\frac{b(c+1)}{1+b}\geq \frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}$

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#7 h Apr 21, 2007, 3:23 AM • 3 Y

Y by ywq233, Adventure10, Mango247

Another solution for the problem:
Let $P$ be $\sum_{cyc}\frac{1}{a(1+b)}$
Then, apply the well-known inequality: $(x+y+z)^{2}\geq 3(xy+yz+zx)$ we obtain:
$P^{2}\geq 3( \sum_{cyc}\frac{1}{ab(1+b)(1+c)})= \frac{3}{abc}-\frac{3}{(1+a)(1+b)(1+c)}-\frac{1}{abc((1+a)(1+b)(1+c)}$
Denote $t= \sqrt[3]{abc}$ and apply AM-GM, we have:
$(1+a)(1+b)(1+c) \geq (t+1)^{3}$
$\Rightarrow P^{2}\geq \frac{3}{t^{3}}-\frac{3}{(t+1)^{3}}-\frac{3}{t^{3}(t+1)^{3}}= \frac{9}{t^{2}(t+1)^{2}}$
Thus
$\sum_{cyc}\frac{1}{a(1+b)}\geq \frac{3}{\sqrt[3]{abc}(\sqrt[3]{abc}+1)}$

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214 posts

shalex

#8 h Apr 21, 2007, 5:05 AM • 3 Y

Y by Arshia.esl, Adventure10, Mango247

I think we can also kill it by a kind of natural method.

Take $a=\lambda x/y,\ b=\lambda y/z, c=\lambda z/x\ (x,y,z,\lambda>0)$ we just need to prove
$\frac{yz}{\lambda xy+zx}+\frac{zx}{\lambda yz+xy}+\frac{xy}{\lambda zx+yz}\geqslant\frac{3}{\lambda+1}$
Take $u=yz,\ v=zx,\ w=xy$ and it comes down to
$\frac{u}{\lambda w+v}+\frac{v}{\lambda u+w}+\frac{w}{\lambda v+u}\geqslant\frac{3}{\lambda+1}$
Use Cauchy-Schwarz we have
$\left(\frac{u}{\lambda w+v}+\frac{v}{\lambda u+w}+\frac{w}{\lambda v+u}\right)[u(\lambda w+v)+v(\lambda u+w)+w(\lambda v+u)]\\ \geqslant(u+v+w)^{2}$
And notice that
$u(\lambda w+v)+v(\lambda u+w)+w(\lambda v+u)=(\lambda+1)(uv+vw+wu)\leqslant\frac{\lambda+1}{3}(u+v+w)^{2}$
So we get what we need.

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827 posts

#9 h May 6, 2007, 9:34 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Another solution:

We substitute $a$ , $b$ , $c$ for $a^{3}$ , $b^{3}$ , $c^{3}$ , so the problem becomes:

Quote:

Prove that
$\frac{1}{a^{3}(1+b^{3})}+\frac{1}{b^{3}(1+c^{3})}+\frac{1}{c^{3}(1+a^{3})}\geq\frac{3}{abc(1+abc)}$
for positive real numbers $a,b,c$ .

$\frac{1}{a^{3}(1+b^{3})}+\frac{1}{b^{3}(1+c^{3})}+\frac{1}{c^{3}(1+a^{3})}-\frac{3}{abc(1+abc)}$

$\equiv F(a,b,c)=F\left(\sqrt[3]{\frac{ky}{x}},\sqrt[3]{\frac{kz}{y}},\sqrt[3]{\frac{kx}{z}}\right)\equiv\frac{G(x,y,z)}{(k^{2}+k)(x+ky)(y+kz)(z+kx)}$ .

Without losing generality, we may assuming $x=\min\{x,y,z\}$ .

$G(x,y,z)=G(x,x+s,x+t)$

$=(k+1)(k^{2}+1)(s^{2}-st+t^{2})x$

$+(s+kt)[\sqrt{k}(2k-3\sqrt{k}+2)st+(k+1)(\sqrt{k}s-t)^{2}]\geq 0$ .

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