Baltic Way Part 1
by djmathman, Nov 13, 2014, 2:54 AM
Problem 1 wrote:
Show that
![\[\cos(56^\circ)\cdot\cos(2\cdot 56^\circ)\cdot\cos(2^2\cdot 56^\circ)\cdot\ldots\cdot\cos(2^{23}\cdot 56^\circ)=\dfrac1{2^{24}}.\]](//latex.artofproblemsolving.com/0/5/4/054457d8caba8b45c6a890903ae115104fbd0b1e.png)
![\[\cos(56^\circ)\cdot\cos(2\cdot 56^\circ)\cdot\cos(2^2\cdot 56^\circ)\cdot\ldots\cdot\cos(2^{23}\cdot 56^\circ)=\dfrac1{2^{24}}.\]](http://latex.artofproblemsolving.com/0/5/4/054457d8caba8b45c6a890903ae115104fbd0b1e.png)
,
,
,
Multiplying all twenty-four equalities together and setting 
yields![\[\cos(56^\circ)\cdot\cos(2\cdot 56^\circ)\cdot\cos(2^2\cdot 56^\circ)\cdot\ldots\cdot\cos(2^{23}\cdot 56^\circ)=\dfrac{\sin(2^{24}56^\circ)}{2^{24}\sin 56^\circ}.\]](http://latex.artofproblemsolving.com/f/1/2/f12e1d2b7c8d06905cf7c07a1a2a47e89dd3b3cf.png)
,
as 
,
.
so it suffices mod 
,
,
.
,
,
.
satisfies all of these properties, we get 
as desired. 
Problem 2 wrote:
Let 
be real numbers satisfying 
and
![\[a_{i+1}-2a_i+a_{i-1}=a_i^2\]](//latex.artofproblemsolving.com/9/0/c/90c7bd11382a1e29166b61667600f8a377ec469b.png)
for
. Prove that 
for .
be real numbers satisfying 
and![\[a_{i+1}-2a_i+a_{i-1}=a_i^2\]](http://latex.artofproblemsolving.com/9/0/c/90c7bd11382a1e29166b61667600f8a377ec469b.png)
for
. Prove that 
for .
,
for each 
.
we get 
,
is a nondecreasing sequence. Next, from the definition of the 
we obtain![\[b_0+b_1+b_2+\cdots+b_{N-1}=a_N-a_0=0.\]](http://latex.artofproblemsolving.com/e/e/1/ee116aa3771d5b9d3a7dbb2ec6e931b66593e1a5.png)
then 
for each 
,
.
,
.
is positive, then there must exist some 
.
for all 
.Problem 3 wrote:
Positive real numbers 
, 
, 
satisfy 
. Prove the inequality
![\[\dfrac1{\sqrt{a^3+b}}+\dfrac1{\sqrt{b^3+c}}+\dfrac1{\sqrt{c^3+a}}\leq\dfrac3{\sqrt2}.\]](//latex.artofproblemsolving.com/c/8/0/c8010f1d911b733b47ad6550f29920d688bccc90.png)
, 
, 
satisfy 
. Prove the inequality![\[\dfrac1{\sqrt{a^3+b}}+\dfrac1{\sqrt{b^3+c}}+\dfrac1{\sqrt{c^3+a}}\leq\dfrac3{\sqrt2}.\]](http://latex.artofproblemsolving.com/c/8/0/c8010f1d911b733b47ad6550f29920d688bccc90.png)
and 
be positive real numbers such that 
and ![$m^3n=\tfrac{\sqrt[3]{2}}{81}$](http://latex.artofproblemsolving.com/4/2/b/42b9dd5d486ffba2e947618a0b22b76e241187d9.png)
.![\[\dfrac1{\sqrt{a^3+b}}\leq\dfrac ma+\dfrac nb.\]](http://latex.artofproblemsolving.com/6/6/6/666802069fffad8f86e74310bd801d021590d34c.png)
denote this above expression. Now note that assigning "weights" of 
to the 
and 
terms and 
to everything else gives an AM-GM mess of![\begin{align*}E&\geq12\sqrt[12]{(n^2a^5)(2mna^4)\left(\dfrac14m^2a^3b^2\right)^4(n^2a^2b)(2mnab^2)\left(\dfrac14m^2b^3\right)^4}\\&=12\sqrt[12]{\dfrac{n^6m^{18}a^{24}b^{24}}{2^{14}}}=a^2b^2\left(\dfrac{9\sqrt{nm^3}}{2^{1/6}}\right)=(ab)^2\end{align*}](http://latex.artofproblemsolving.com/c/9/f/c9f5a2d35612d619aa4a212c76e01418ede18b5e.png)
![\[\sum\dfrac1{\sqrt{a^3+b}}\leq\sum\left(\dfrac ma+\dfrac nb\right)=(m+n)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)=\dfrac{3}{\sqrt2}.\qquad\blacksquare\]](http://latex.artofproblemsolving.com/a/d/5/ad5d1b88416f7d7dec7f0bc12200d8d4d5c5828b.png)
This post has been edited 2 times. Last edited by djmathman, Mar 5, 2015, 3:43 AM
Reason: align tags
Reason: align tags
can't you just do![$ \sum_{cyc}\frac{1}{\sqrt{a^3 + b}} \le \frac{1}{\sqrt{2}}\sum_{cyc}\frac{1}{\sqrt[4]{a^3b}} \le \frac{1}{4\sqrt{2}}\sum_{cyc}\frac{3}{a} + \frac{1}{b} = \frac{3}{\sqrt{2}} $](http://latex.artofproblemsolving.com/6/7/b/67ba4ee0995b389083daed9b90fd96a6a3834ba3.png)
by two applications of AM-GM (the second one is weighted)? Lol #overkill 
April 2025
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