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be positive real numbers such that ![\(a+b+c=4\sqrt[3]{abc}\)](http://latex.artofproblemsolving.com/2/8/2/28214a1814762b71858db1543fae1a50212eaafa.png)
. Prove that ![\[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]](http://latex.artofproblemsolving.com/7/0/4/7043a170ac5535465a5a1b4daccb1f1de3043280.png)
+2 w
, and it is easy to see that this works.
, 
, 
, and 
. Putting these into the equation gives us![\[f(a)+f(a+3d)=f(a+d)+f(a+2d)\]](http://latex.artofproblemsolving.com/a/f/3/af31cd2951970fb7ab3213a9a3048ec11b6b1244.png)
If we subtract 
from everything, this simplifies to![\[f(a-d)+f(a+2d)=f(a)+f(a+d)\]](http://latex.artofproblemsolving.com/6/9/e/69e61a43fc63041853e3d9fcc4be10686d517d0a.png)
Then, if we add the two equations together and cancel, we get![\[f(a-d)+f(a+3d)=2f(a+d)\]](http://latex.artofproblemsolving.com/5/0/a/50aee1955fc8ff30cc56ead95a6ebf621260c958.png)
This is Jensen's Functional Inequality, which we know has solution , but only if 
. So if 
, then 
, but this contradicts the fact that 
, so can't be 
, so we're done. 
denote the given assertion and be some nonzero rational number.
.
.
.
and can be any rational numbers, this simplifies to 
. This is Jensen's FE, but I will solve it here.
. Then 
with the additional constraint that 
. Let 
be this new assertion.
.
, so 
is Cauchy, which is well-known to have solutions 
for some constant 
over the rationals.
, 
for rational constants and 
. To show that this works, we have 
. Note if 
, then constant solution obviously works. If not, then 
, which is true since they form an arithmetic progression.
for all , where 
. These work since![\[ (ax + b) + (at + b) = (ay + b) + (az + b)\iff a(x + t - y -z) = 0,\]](http://latex.artofproblemsolving.com/c/b/c/cbc17dd7de1beecae87c00ffb48c3a171ed72571.png)
which is true since 
. It remains to show that these solutions are the only ones.
and 
separately to get the equations![\[ f(a) + f(a + 3d/2) = f(a + d/2) + f(a + d)\qquad (1),\]](http://latex.artofproblemsolving.com/7/c/5/7c58ddc87288a4b36af54e7c4b16b0b43f2e2405.png)
![\[ f(a + d) + f( a + 3d/2) = f(a +d/2) + f(2d)\qquad (2).\]](http://latex.artofproblemsolving.com/d/9/1/d91c76a113d86df3f6ac2fa3d1b1a8e5a812801f.png)
Subtracting 
from 
gives 
. Using the substitution 
, we have 
for all 
. Hence, 
satisfies Jensen's FE over the rationals, so is linear.
, or 
for rational 
and 
, which works. Now, we show these are the only solutions.
be in arithmetic progression. Then
Subtracting, we get 
, so is linear over all arithmetic progressions in 
. In particular, is linear for all integer inputs. Let be a linear function over such that 
for all integral . We will show that 
.
be a nonzero rational number, where 
and 
are nonzero integers. Since is linear over 
, and 
and 
, we know that 
over that set, so 
. Hence, , as desired.
and 
. Then, 
. We can plug in 
for and we get 
. We add these equation to get 
. We can plug in 
for to get 
. We can plug in 
for to get 
for all rationals and . It is easy to see by induction that 
for integers 
. This means that 
so 
so 
for all rational . Plugging this back in, all 
work.
. We may now consider the quantities 
for some integer and positive rational ; by induction we find that the values 
actually form an arithmetic sequence. This immediately solves the problem over 
and subsequently . 
and 
,
Thus we have 
so satisfies Jensen's functional equation. As a result, 
for constants are the only solutions.
work.
we have
So we get 
, equivalent to
Doing induction similar to the Cauchy Functional Equation over gives the solution set.
is of the form 
for some 
, easy to check why they work and we shall show that these are the only solutions to .
, with 
and . Plug these terms in the following manner,
Adding two of the above equation we deduce to the following,
Aha! this resembles the ubiquitous Jensen's equation and turns out it is precisely that as one may substitute 
and 
( this is because 
can generate the whole of 
). 
such that![\[f(x)+f(t)=f(y)+f(z)\]](http://latex.artofproblemsolving.com/7/e/c/7ecb395c1e56e6d9726fc45f1bc7a66d98333da2.png)
for all rational numbers 
for any rational a and b.
.
so that 
.
.
.
and 
for 
,
is linear, and this is true for all 
and 
now add
for integer 
,
with 
where 
to get ![\[f(a) + f(a+3d) = f(a+d) + f(a+2d).\]](http://latex.artofproblemsolving.com/8/a/0/8a08b0fc1e6757479f8921e9b1af170894293075.png)
Now replace 
to get ![\[f(a-d) + f(a+2d) = f(a) + f(a+d).\]](http://latex.artofproblemsolving.com/9/d/4/9d4462cbe7d7333d382eda216e50aa829dd616aa.png)
We can add these two equations so that terms cancel: ![\[f(a) + f(a+3d) + f(a-d) + f(a+2d) = f(a+d) + f(a+2d) + f(a) + f(a+d),\]](http://latex.artofproblemsolving.com/1/e/0/1e0d005858feb5313d499480d2cc6cf0a5ed7d62.png)
which simplifies to ![\[f(a-d) + f(a+3d) = 2f(a+d).\]](http://latex.artofproblemsolving.com/5/b/5/5b55ed29090e5ea1aabe2682ddeb5c6bbac560c9.png)
For 
this is equivalent to ![\[f(m) + f(n) = 2f\left(\dfrac{m+n}{2}\right).\]](http://latex.artofproblemsolving.com/5/d/3/5d3f17d1c1381f9b95f19bc1f094dd71c4e1175d.png)
(This is Jensen's functional equation, by the way.) I wanted to reduce this to Cauchy's functional equation somehow, which I believe is citable, but I'm still not too sure, so I set 
and 
to get 
Hence ![\[f(m) + f(n) = f(0) + f(m+n).\]](http://latex.artofproblemsolving.com/9/7/8/978db954bf33cfc7283a938021770eaf61bbc5d7.png)
Now just let 
and substitute to get ![\[g(x) + g(y) = g(x+y).\]](http://latex.artofproblemsolving.com/1/2/f/12f03cce4f287b98b81e48d682544411467efef5.png)
This is Cauchy's functional equation over the rationals, hence 
for some 
so 
for some 
.
.
,
.
,
works, so that is our answer.
my solution was wrong 
which satisfies.
be the given assertion. Just compare 
and 
to get 
as desired.
and 
.
which implies (after scaling back) that any arithmetic progression's image is also an arithmetic progression; this immediately finishes.
This is a four-term equation, so clearly something must cancel - we can rewrite as 
Adding these two and simplifying gives 
but this is just Jensen's FE, which has solutions at 
for rationals 
.