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such that for all real 
![\[f(x+f(y))=f(x+y)+y\]](http://latex.artofproblemsolving.com/1/e/2/1e2fef35ee53cd3224527f672b9edbb7a5293fad.png)
and 
such that 
divides 
and 
divides .
and two points 
and 
outside , a quadrilateral 
is defined as "good" if 
are four distinct points on in order, and lines 
and 
intersect at and lines 
and 
intersect at .
, let 
denote its area. If there exists a good quadrilateral, prove that there exists good quadrilateral such that for any good quadrilateral 
, 
.
Prove that the number of divisors of the form 
of each positive integer is not less than the number of its divisors of the form 
, 
, 
be positive real numbers such that 
. Prove that![\[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
\]](http://latex.artofproblemsolving.com/b/2/9/b295b0b2a765e41a639657e13b724c60323fa208.png)
![\[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \Big( a(b+c)+b(c+a)+c(a+b) \Big) \geq \left( \frac 1a +\frac 1b+ \frac 1c \right)^2 \]](http://latex.artofproblemsolving.com/d/d/e/ddec9f4173c17ad4397ac49b9b124f7598f5e986.png)
and as 
we obtain that ![\[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \geq \frac{ ab+bc+ca} 2 \geq \frac{\sqrt [3]{a^2b^2c^2}}{2} = \frac 32 \]](http://latex.artofproblemsolving.com/f/9/2/f92584373ccb6c03721548eb6389ccae6cf7a2f6.png)
which is what we wanted to prove.
i came to a weird solution, but cool and nice!
such that
.
and use that 
, then we get to
or 
or 
.
, so we want 
, this is
, so we want 
, or 
.
.
and let 
and 
.
or (after a bit of simplifications)
and we're done!
and analogically for 
and 
, so the assumption that 
is satisfied, ok
then we prove it pretty same as Nesbitt
be positive real numbers such that 
.![\[ \frac{\sqrt [3]{a^{2}b^{2}c^{2}}}{2}= \frac{3}{2}\]](http://latex.artofproblemsolving.com/a/a/1/aa12ff3fab5b3174e2c9c440d816c7a7dda442fa.png)
?
be positive real numbers such that .?![${ \frac{3}{2}\sqrt [3]{a^{2}b^{2}c^{2}}}= \frac{3}{2}$](http://latex.artofproblemsolving.com/e/d/c/edc3a16ce082e110fa2cd8d184a74aa7f483b8cd.png)
.
in some form and know that the expression 
would similarly give you this expression?
but 
the left hand side, etc.) and try applying them until something worked?
with 
. We still have 
. We aim to prove 
or equivalently 
.
.![$\sum\frac{x^2}{y+z}\ge \frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}$](http://latex.artofproblemsolving.com/d/c/9/dc9046bc151221668a4a4fff717e2b2973c748b0.png)
.
.
with . We still have . We aim to prove or equivalently .
, , be positive real numbers such that . Prove that
, 
and 
. Inequality becomes: 
.
inequality we have: 
![$LHS\geq \frac{z+y+x}{2}\geq \frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}=RHS$](http://latex.artofproblemsolving.com/f/3/2/f32145d673e1a8171702464e356f28824b4fffcd.png)
. So we are done:)
![$\sum bc \geq 3\sqrt[3]{(abc)^{2}}=3$](http://latex.artofproblemsolving.com/c/c/7/cc7ed6030cb2ed75e60bceee3a189ffdbb43d12c.png)
by ![\[\sum \frac{1}{a^{3}(b+c)}=\sum \frac{b^{2}c^{2}}{a^{3}b^{2}c^{2}(b+c)}=\sum\frac{(bc)^{2}}{ab+ac}\geq \frac{\left(\sum bc \right)^{2}}{2\sum bc}=\frac{\sum bc}{2}\geq \frac32\]](http://latex.artofproblemsolving.com/c/a/c/cac6a0cd36d68da41416e3328656ea1a3da47a5f.png)
as desired.
by Titu's lemma and AM-GM respectively.
![\[\left(\sum_{\text{cyc}} a(b+c)\right)^{\frac{1}{2}} \left(\sum_{\text{cyc}} \frac{1}{a^3(b+c)} \right)^{\frac{1}{2}} \geq \left(\frac 1a + \frac 1b + \frac 1c \right) = (ab+ac+bc)\]](http://latex.artofproblemsolving.com/0/2/e/02e8f6df91c9612bf38ae576d09311051b902d40.png)
Thus it remains to prove that 
.
works
From Muirhead's Inequality we have
Adding them all together gives the desired inequality.
As 
M
on the 
,
[By Titu's Lemma]![$\frac{ab+bc+ca}{2} \geq \frac{3 \sqrt[3]{a^2b^2c^2}}{2} = \frac{3}{2}$](http://latex.artofproblemsolving.com/3/4/b/34b679ad0c98c7754b5c7b47e8ab890ca0a2ffb4.png)
by AM-GM. Hence, we're done!
then
.
By Titu's Lemma,
By AM-GM, note that ![$ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2} = 3$](http://latex.artofproblemsolving.com/9/6/3/96305d02b6d624b4f41449e862b62ffe93fb3a4f.png)
,
![$\frac{ab+bc+ac}{2*3} \ge \frac{\sqrt[3]{(abc)^2)}}{2}$](http://latex.artofproblemsolving.com/9/c/e/9ce0de401a7c0a705075ea57261cac992c548afd.png)
![$\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{ab+bc+ac}{2}\ge \frac{\sqrt[3]{(abc)^2}*3}{2}\ge \frac{3}{2}$](http://latex.artofproblemsolving.com/f/c/d/fcd403739a6089f6ce2ce3aa20d810a9ba9c5293.png)
This inequality is by Cauchy-Schwarz. It is sufficient to prove that
![$\sum\frac{1}{a^3(b+c)}=\sum\frac{\frac1{a^2}}{ab+ac}\geq\frac{(\frac1{a}+\frac1{b}+\frac1{c})^2}{2(ab+ac+bc)}=\frac{(ab+bc+ca)^2}{2(ab+ac+bc)}=\frac{(ab+bc+ca)}{2}\geq\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac32$](http://latex.artofproblemsolving.com/2/a/6/2a64bee273403dff6e2ebd04590664309f773b3b.png)
,
,
.![\[\sum_{\text{cyc}} \frac{1}{\frac{1}{x^3}\left(\frac{1}{y} + \frac{1}{z} \right)} = \sum_{\text{cyc}} \frac{1}{\frac{y + z}{x^3yz}} = \sum_{\text{cyc}} \frac{1}{\frac{y + z}{x^2}} = \sum_{\text{cyc}} \frac{x^2}{y+z}\]](http://latex.artofproblemsolving.com/d/5/1/d511ec8666c175d63e4b0153f3e39ed43e60ad21.png)
![\[\sum_{\text{cyc}} \frac{x^2}{y+z} \geq \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2} \geq \frac{3\sqrt[3]{xyz}}{2} = \frac{3}{2}\]](http://latex.artofproblemsolving.com/5/1/d/51dd548779258039e7bc5740ccfa14964c972f65.png)
