[/quote]
,
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
,
,
be nonzero real numbers such that 
and 
.
.
kawaii if it satisfies 
is the number of integers not more than 
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers 
and 
.
and 
for all integers 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
,
,
is a multiple of 
.
a point in the space and 
the centroid of a tetrahedron 
.
be a 
on it, line 
makes a fixed angle with the tangent to 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
on the spiral, the polar of 
be a monic polynomial so that there exists another real coefficients 
that satisfy![\[P(x^2-2)=P(x)Q(x)\]](http://latex.artofproblemsolving.com/6/9/7/697faa929e4fda7a6e9b1cd97849bd42ffc14306.png)
Determine all complex roots that are possible from 
with exactly 9 elements that is subset of 
.
and 
that satisfy the following:
is an empty subset.
is 
and 
.
to 
is 
and 
.
and 
is 
is 
is tangent to the 
Y by Amir Hossein, rightways, Davi-8191, Mathuzb, Rogan128, e_plus_pi, Wizard_32, Myriam2003, mathleticguyyy, Lcz, Loveineq., megarnie, centslordm, Lamboreghini, GoodMorning, Adventure10, Mango247, cubres
Let 
, 
, 
be positive real numbers such that 
. Prove that ![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](//latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
Proposed by Tigran Margaryan, Armenia
, 
, 
be positive real numbers such that 
. Prove that ![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
Proposed by Tigran Margaryan, Armenia
This post has been edited 1 time. Last edited by zsalomon, Dec 13, 2022, 6:43 PM
Reason: proposal credit added
Reason: proposal credit added
and 
then ![\[ (x^2+1)(y^2+1) \le \left( \left( \frac{x+y}{2} \right)^2+1 \right)^2. \]](http://latex.artofproblemsolving.com/9/a/4/9a4298bbe7986afaf7174434e99cce85176ceeb4.png)
![\[ \frac1{16} (x-y)^2 \left( x^2+6xy+y^2-8 \right) \]](http://latex.artofproblemsolving.com/e/3/b/e3b1611f106c9476a883e25acfe8799155d2ba8f.png)
which is nonnegative. 
,
,
where 
and 
also factors out.
we get 
,
.
of positive reals valid, if 
.
.
we have 
by AM-GM, hence the claim is true. 
is at least 
by AM-GM inequality, where 
and 
.
for all 
.
for all valid triples 
is the second derivative of 
,
.
such that 
(Proof) Observe that this is equivalent to 
whenever 
which is obviously true now. 
The last inequality holds by Jensen's on 
for any reals 
with 
.
so we have
and 
and 
since from AM-GM we have
and 
and 
hence
for 
for 
we have
![$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)}\leq \left(\frac{a+b+c}{3}\right)^2+1$](http://latex.artofproblemsolving.com/0/6/1/061ca9b80f24cd7487d358d9940b5ff87cc8f253.png)
be positive reals such that the product of any two distinct 
is at least 1. Prove that![\[
\prod_{i=1}^{n} (a_i^2+1) \leq \left( \sum_{1\leq i<j\leq n} \frac{2a_ia_j}{n(n-1)}+1 \right)^{n-1} \left( \sum_{i=1}^{n} \frac{a_i^2}{n}+1 \right)
\]](http://latex.artofproblemsolving.com/1/2/5/125c2346f5d3ae6787412b05451d869c54c71f01.png)
.
.
,![\[ f\left(\frac{a+b+c}3\right)=f\left(\frac{x+y+z}3\right)\ge\frac{f(x)+f(y)+f(z)}3, \]](http://latex.artofproblemsolving.com/f/d/d/fdd920bf52cef7dfd098253ea322fe21ced16c6e.png)
so showing 
is sufficient, which is equivalent to ![\[ (a^2+1)(b^2+1) \le \left( \left( \frac{a+b}{2} \right)^2+1 \right)^2. \]](http://latex.artofproblemsolving.com/2/0/4/2044e1191eae4e3691b29322d4a6c94641a9aedd.png)
But subtracting the LHS from the RHS produces ![\[ \frac1{16} (a-b)^2 \left( a^2+6ab+b^2-8 \right), \]](http://latex.artofproblemsolving.com/0/a/e/0aeb91858f8ce04fe1134809635fd18b7c40327f.png)
which is clearly nonnegative.
,
.
)
with 
yields a greater LHS (which can be proven by rearrangement and factoring), as desired.
then 
.
expands to 
which is obviously true.
which is concave on ![\[ f(a) + f(b) + f(c) \le 3f\left(\frac{a+b+c}{3}\right). \]](http://latex.artofproblemsolving.com/a/b/c/abce7de94f9126018928d899b8c885005833b9b3.png)
,![\[ \sqrt{(a^2+1)(b^2+1)} \le \left( \frac{a+b}{2} \right)^2 + 1. \]](http://latex.artofproblemsolving.com/3/3/5/3356fb1ecf375e9e2dbfcd153af7074249963a60.png)
Proof. This expands as 
which follows as 
.![\[ f(a) + f(b) + f(c) \le 2f\left(\frac{a+b}{2}\right) + f(c) \le 3\left(\frac{a+b+c}{3}\right) \]](http://latex.artofproblemsolving.com/1/e/5/1e525471c5525616cd740fdebcc96c3d2f40b51c.png)
as 
.
![\[ (x^2 + 1)(y^2 + 1) \le \left( \left( \frac{x + y}{2} \right)^2 + 1 \right)^2 \]](http://latex.artofproblemsolving.com/4/8/c/48c22467c84dd7a5baa5a0569238cafba4621108.png)
becomes![\[ \frac 1{16} (x-y)^2(x^2 + 6xy + y^2 - 8) \ge 0 \]](http://latex.artofproblemsolving.com/b/3/4/b344ae3afb6e0fc31f29d434edca319be0034fb2.png)
which is true because 
.![\[ \frac{(a + b)(c + d)}{4} \ge 1 \iff ab + bc + cd + da \ge 4, \]](http://latex.artofproblemsolving.com/f/6/d/f6d8070967371393340477b9a8f511a9f63dd371.png)
which is true by assuming 
.![\[ \sqrt[4]{(a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1)} \le \left( \frac{a + b + c + d}{4} \right)^2 + 1. \]](http://latex.artofproblemsolving.com/8/2/d/82dde22eb17d72a2b2d5dffff37ef000a803146c.png)
Setting 
,
,
,
and 
,![\[ (a^2 + 1)(b^2 + 1)(c^2 + 1)\left( \left(\frac{a + b + c}{3} \right)^2 + 1 \right) \le \left( \left( \frac{a + b + c}{3} \right)^2 + 1 \right)^4 \]](http://latex.artofproblemsolving.com/5/8/2/5824333b6d405115cfff2b719ee89ec9c63b34ca.png)
which rearranges into the result.
.
,![\[\frac{f(a)+f(b)+f(c)}{3}\leq f\left(\frac{a+b+c}{3}\right).\]](http://latex.artofproblemsolving.com/0/4/8/0482b78d9bb42148d45911840a5a70ddda309411.png)
This is true by Jensen's inequality on the interval 
,
.
.![\[f(a)+f(b)+f(c)\leq f(x)+f(y)+f(c),\]](http://latex.artofproblemsolving.com/1/3/9/139ed62a60bf8c70f6bec1b0966ddcf769605a29.png)
then:![\[f(a)+f(b)+f(c)\leq 3f\left(\frac{a+b+c}{3}\right).\]](http://latex.artofproblemsolving.com/6/8/3/68322004f3086975d3b234cb7a8b1208e50b1d77.png)
![\[\sqrt{(a^2+1)(b^2+1)}\leq \left(\frac{a+b}{2}\right)^2+1,\]](http://latex.artofproblemsolving.com/9/9/a/99a8252e3b32acde512ea42bb6cc0787685cd4d3.png)
![\[16(a^2b^2+a^2+b^2+1)\leq (a+b)^4+16+8(a+b)^2,\]](http://latex.artofproblemsolving.com/8/a/c/8ac2700a8d409a5a82ef8f26e7a9a592922fe813.png)
![\[(a+b)^4-16a^2b^2\geq 8(a-b)^2,\]](http://latex.artofproblemsolving.com/7/9/5/795d554f33828ca09361ed721b24a973c834349f.png)
![\[(a-b)^2(a^2+6ab+b^2)\geq 8(a-b)^2,\]](http://latex.artofproblemsolving.com/2/c/6/2c6c3cce2e95acc53331305277934514ed039422.png)
which is true by AM-GM, so we are done.
be positive reals 
)
.![$$\sqrt[2^p-x]{\prod{a_{1}^2+\lambda}}\leq \left(\dfrac{\sum_{cyc}{a_{1}}}{2^p-\lambda}\right)^2+\lambda$$](http://latex.artofproblemsolving.com/b/d/c/bdc7cd73e3d5a8a319d0b1a3095925722c9c18f9.png)
Then 
so if 
then 
and 
so Jensen's gives ![\[\frac{f(a)+f(b)+f(c)}3\le \frac23f\left(\frac{a+b}2\right)+\frac13f(c)\le f\left(\frac{a+b+c}3\right),\]](http://latex.artofproblemsolving.com/2/a/8/2a801f288b11129e1e840cd127781652d1ddfbde.png)
and taking 
to the power of both sides gives the desired result.
,
,
be positive real numbers such that: ![$\min \{ a_ia_j \mid 1 \leq i<j\leq n \} ] \ge k$](http://latex.artofproblemsolving.com/7/0/4/7045c9129db4c811f034aca9fa443b0e89f141ac.png)
.![$$\sqrt[n]{\prod_{i=1}^{n}(a_i^2+k)} \le \left(\frac{\sum_{i=1}^{n}a_i}{n}\right)^2 + k.$$](http://latex.artofproblemsolving.com/d/0/a/d0a0eddfea89b8b9be80b62111c9484c5a54a49c.png)
in there isn't a generalization. Substitute 
so 
and you get exact same output with the original inequality which doesn't generalize it
Substititue 
and 
and 
.
,![$\sqrt[3] {(a^2 + 1)(b^2 + 1)(c^2 + 1)}$](http://latex.artofproblemsolving.com/5/7/b/57bdf1f1c0244bb89343a25de29dc7333e4d8848.png)
.
,
which is further equivalent to 
which is further equivalent to 
,
,
and our condition.
is nondecreasing by this operation. To see this, without loss of generality let 
.
increases and 
goes from 
to 
.
satisfies the conditions of the problem and has ![$x + y + z = c, k = \sqrt[3]{\prod_{cyc} x^2 + 1} > (\frac c3)^2 + 1 $](http://latex.artofproblemsolving.com/b/9/f/b9fa38eb5d931de53a8e7017d34d0d1cdb33ad28.png)
.
.
such that the maximum distance from 
that one of the numbers is is less than some number ![$\frac{\sqrt[3]{((p^2 + 1) (q^2 + 1)(r^2 + 1))}}{(\frac c3)^2 + 1} \ge m$](http://latex.artofproblemsolving.com/7/e/f/7ef0f7bf849b3eeebbca67b7791e60ab05877aa1.png)
.
is 
which is a polynomial function of ![$m \le \sqrt[3] {g^3} = g$](http://latex.artofproblemsolving.com/6/6/d/66d5cc425ff91c60534e4c7744757e7eb16f072a.png)
,
But observe that the function 
so the above inequality holds if 
so WLOG we must have 
Now the following lemma, for 
holds: 
The proof is straight up expansion and then simplifying.
where the last inequality holds by the same Jensen's as above because 
by AM-GM. QED
