Let x1,x2,…,x2014x_1,x_2,\ldots ,x_{2014}x1,x2,…,x2014 be positive real numbers such that ∑j=12014xj=1\large\displaystyle\sum^{2014}_{j=1} x_j=1j=1∑2014xj=1 Determine with proof the smallest constant KKK such that K∑j=12014xj21−xj≥1\large K \displaystyle\sum^{2014}_{j=1} \dfrac{x^2_j}{1-x_j}\geq1Kj=1∑20141−xjxj2≥1
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Note by Aneesh Kundu 6 years, 6 months ago
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x1+x2+…+x2014=1 x_{1} + x_{2} + \ldots + x_{2014} = 1x1+x2+…+x2014=1
x121−x12+x221−x22+...........................+x201421−x20142\dfrac{x_{1}^{2}}{1 - x_{1}^{2}} + \dfrac{x_{2}^{2}}{1 - x_{2}^{2}} + ........................... + \dfrac{x_{201}4^{2}}{1 - x_{2014}^{2}}1−x12x12+1−x22x22+...........................+1−x20142x20142
Applying Titu's Lemma ,
x121−x12+x221−x22+...........................+x201421−x20142≥(x1+x2+....+x2014)22014−(x1+x2+.....+x2014)\dfrac{x_{1}^{2}}{1 - x_{1}^{2}} + \dfrac{x_{2}^{2}}{1 - x_{2}^{2}} + ........................... + \dfrac{x_{201}4^{2}}{1 - x_{2014}^{2}} \geq \dfrac{(x_{1} + x_{2} + .... + x_{2014})^{2}}{2014 - (x_{1} + x_{2} + .....+ x_{2014})}1−x12x12+1−x22x22+...........................+1−x20142x20142≥2014−(x1+x2+.....+x2014)(x1+x2+....+x2014)2
x121−x12+x221−x22+...........................+x201421−x20142≥12013\dfrac{x_{1}^{2}}{1 - x_{1}^{2}} + \dfrac{x_{2}^{2}}{1 - x_{2}^{2}} + ........................... + \dfrac{x_{201}4^{2}}{1 - x_{2014}^{2}} \geq \frac{1}{2013}1−x12x12+1−x22x22+...........................+1−x20142x20142≥20131
Thus , K=2013 \boxed{K = 2013}K=2013
Method 2
∑12014xj2−11−xj+11−xj \displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j} - 1}{1 - x_{j}} + \dfrac{1}{1 - x_{j}}1∑20141−xjxj2−1+1−xj1
∑12014−(1+xj)+11−xj \displaystyle \sum_{1}^{2014} -( 1 + x_{j}) + \dfrac{1}{1 - x_{j}}1∑2014−(1+xj)+1−xj1
−2015+∑1201411−xj=∑12014xj21−xj \boxed{- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} = \displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j}}{1 - x_{j}}}−2015+1∑20141−xj1=1∑20141−xjxj2
Applying A.M≥H.M A.M \geq H.MA.M≥H.M
∑1201411−xj2014≥20142014−(x1+x2.....x2014)\frac{\displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}}}{2014} \geq \dfrac{2014}{ 2014 - ( x_{1} + x_{2} ..... x_{2014})}20141∑20141−xj1≥2014−(x1+x2.....x2014)2014
∑1201411−xj≥201422013\displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2}}{ 2013}1∑20141−xj1≥201320142
−2015+∑1201411−xj≥201422013−2015 - 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2}}{ 2013} - 2015−2015+1∑20141−xj1≥201320142−2015
−2015+∑1201411−xj≥20142−2015×20132013 - 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2}- 2015 \times 2013}{2013} −2015+1∑20141−xj1≥201320142−2015×2013
−2015+∑1201411−xj≥20142−(2014+1)(2014−1)2013 - 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2} - (2014 + 1)(2014 - 1)}{2013} −2015+1∑20141−xj1≥201320142−(2014+1)(2014−1)
−2015+∑1201411−xj≥20142−(20142−1)2013 - 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2} - (2014^{2} - 1)}{2013} −2015+1∑20141−xj1≥201320142−(20142−1)
−2015+∑1201411−xj≥12013 - 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{1}{2013} −2015+1∑20141−xj1≥20131
∑12014xj21−xj≥12013 \displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j}}{1 - x_{j}} \geq \dfrac{1}{2013} 1∑20141−xjxj2≥20131
2013∑12014xj21−xj≥1 2013\displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j}}{1 - x_{j}} \geq 1 20131∑20141−xjxj2≥1
K=2013\boxed{K = 2013}K=2013
Enjoy!!!
\Huge\color{#ff00b3}{♛}\;\;\;\color{#ff0000}{ ❤ }\;\;\;\color{#0000ff}{\mathbf{B}}\color{#ff7f00}{\mathbf{r}}\color{#ffff00}{\mathbf{i}}\color{#00ff00}{\mathbf{l}}\color{#00ffff}{\mathbf{l}}\color{#0000ff}{\mathbf{i}}\color{#8b00ff}{\mathbf{a}}\color{#ff0000}{\mathbf{n}}\color{#ff7f00}{\mathbf{t}}\color{color:#ffff00}{\mathbf{.}}\color{#00ff00}{\mathbf{o}}\color{#0081ff}{\mathbf{r}}\color{#ff69b4}{\mathbf{g}} copied from anastisya romoniva comment
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Your second solution using AM>HM is what I wrote during the RMO, almost verbatim... Wow.
Yes i too enjoyed very much doing by the 2nd method
Yeah!!
Its tagged under A.M G.M inequality , have you done using it , if yes then how
@Sandeep Rathod – Actually I meant AM-HM
Typo
u can use \dfrac(it gives bigger fractions) and also \ldots(it gives the dots)
I applied the \Idots , its not working
@Sandeep Rathod – Its actually small "L" …\ldots…
I know its confusing sometimes
@Aneesh Kundu – Most latex is small letters, unless you want to stress something or make it
More lines: \rightarrow, \RIghtarrow give us →,⇒ \rightarrow , \Rightarrow →,⇒ Capital Greek alphabet: \gamma, \Gamma give us γ,Γ \gamma, \Gammaγ,Γ
Contrary to @Sandeep Rathod solution there is one more method :
Let an function y=x21−xy\quad =\quad \cfrac { \quad { x }^{ 2 } }{ 1-x } \\ y=1−xx2. in (0,1)
So it inscribe an convex polygon of centroid G so from jensons inequality :
y=f(x)=x21−xyG≥yp∑f(xi)n≥f(∑xin)∑f(xi)≥nf(1n)(putn=2014)2013∑f(xi)≥1K=2013y\quad =\quad f(x)\quad =\quad \cfrac { \quad { x }^{ 2 } }{ 1-x } \\ \\ { y }_{ G }\quad \ge \quad { y }_{ p }\\ \\ \frac { \sum { f\left( { x }_{ i } \right) } }{ n } \quad \ge \quad f\left( \cfrac { \sum { { x }_{ i } } }{ n } \right) \\ \\ \sum { f\left( { x }_{ i } \right) } \quad \ge \quad nf\left( \cfrac { 1 }{ n } \right) \quad \quad \quad (put\quad n\quad =\quad 2014)\\ \\ 2013\sum { f\left( { x }_{ i } \right) } \quad \ge \quad 1\\ \\ \boxed { K\quad =\quad 2013 } y=f(x)=1−xx2yG≥ypn∑f(xi)≥f(n∑xi)∑f(xi)≥nf(n1)(putn=2014)2013∑f(xi)≥1K=2013.
Q.E.D
i did'nt understood , how it's inscribed? @Deepanshu Gupta
I appreciate - you always think different
Thanks ! But Your Solution is also very elegant ! I also appreciate it !
And Actually if we consider 'n' points (2014 points) on curve and join them, then they will form an closed loop (or we can say convex polygon ) So Centroid of this Polygon , which is surely lies inside the polygon So it's y-coordinate is greater or equal (when all pt's are collinear , I think) to y-coordinates of point on curve which has same x-coordinate as that of centroid of this loop (Polygon) !
Here I used Jenson's inequality
I got 2013 by Cauchy-Scwartz
Ditto. Do you think solving 4 is enough!Dunno, this waz my only RMO.
I learnt the spelling of "Cauchy Schwartz" today,and I got the solution at my first glance! ;-)
Please post your method @Ranjana Kasangeri
I think there's no 't'.
Yeah! You are right. I got the spelling wrong! * sob * @Joel Tan
Sir,Multiply the LHS by summation or (1-x) ,then apply Cauchy!
i got 201320132013
Anish can you please post the first 5 questions also... This one I did by cauchy Schwartz
Its given below the question
@Aneesh Kundu – Can someone explain the third question please
@Ameya Karode – I did it by contradiction but not very sure..
@Ameya Karode – Is that the one about permuting the digits of two powers of 2?
Chebyshev
Tchebycheff
-- http://en.wikipedia.org/wiki/Chebyshev%27sinequality
I had given my RMO yesterday and done this using Jensen's inequality.
I used Titu's Lemma for this one!
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Method 1
x1+x2+…+x2014=1
1−x12x12+1−x22x22+...........................+1−x20142x20142
Applying Titu's Lemma ,
1−x12x12+1−x22x22+...........................+1−x20142x20142≥2014−(x1+x2+.....+x2014)(x1+x2+....+x2014)2
1−x12x12+1−x22x22+...........................+1−x20142x20142≥20131
Thus , K=2013
Method 2
1∑20141−xjxj2−1+1−xj1
1∑2014−(1+xj)+1−xj1
−2015+1∑20141−xj1=1∑20141−xjxj2
Applying A.M≥H.M
20141∑20141−xj1≥2014−(x1+x2.....x2014)2014
1∑20141−xj1≥201320142
−2015+1∑20141−xj1≥201320142−2015
−2015+1∑20141−xj1≥201320142−2015×2013
−2015+1∑20141−xj1≥201320142−(2014+1)(2014−1)
−2015+1∑20141−xj1≥201320142−(20142−1)
−2015+1∑20141−xj1≥20131
1∑20141−xjxj2≥20131
20131∑20141−xjxj2≥1
K=2013
Enjoy!!!
\Huge\color{#ff00b3}{♛}\;\;\;\color{#ff0000}{ ❤ }\;\;\;\color{#0000ff}{\mathbf{B}}\color{#ff7f00}{\mathbf{r}}\color{#ffff00}{\mathbf{i}}\color{#00ff00}{\mathbf{l}}\color{#00ffff}{\mathbf{l}}\color{#0000ff}{\mathbf{i}}\color{#8b00ff}{\mathbf{a}}\color{#ff0000}{\mathbf{n}}\color{#ff7f00}{\mathbf{t}}\color{color:#ffff00}{\mathbf{.}}\color{#00ff00}{\mathbf{o}}\color{#0081ff}{\mathbf{r}}\color{#ff69b4}{\mathbf{g}} copied from anastisya romoniva comment
Log in to reply
Your second solution using AM>HM is what I wrote during the RMO, almost verbatim... Wow.
Log in to reply
Yes i too enjoyed very much doing by the 2nd method
Yeah!!
Log in to reply
Its tagged under A.M G.M inequality , have you done using it , if yes then how
Log in to reply
Actually I meant AM-HM
Typo
u can use \dfrac(it gives bigger fractions) and also \ldots(it gives the dots)
Log in to reply
I applied the \Idots , its not working
Log in to reply
Its actually small "L" …
I know its confusing sometimes
Log in to reply
Most latex is small letters, unless you want to stress something or make it
More lines: \rightarrow, \RIghtarrow give us →,⇒
Capital Greek alphabet: \gamma, \Gamma give us γ,Γ
Contrary to @Sandeep Rathod solution there is one more method :
Let an function y=1−xx2. in (0,1)
So it inscribe an convex polygon of centroid G so from jensons inequality :
y=f(x)=1−xx2yG≥ypn∑f(xi)≥f(n∑xi)∑f(xi)≥nf(n1)(putn=2014)2013∑f(xi)≥1K=2013.
Q.E.D
Log in to reply
i did'nt understood , how it's inscribed? @Deepanshu Gupta
I appreciate - you always think different
Log in to reply
Thanks ! But Your Solution is also very elegant ! I also appreciate it !
And Actually if we consider 'n' points (2014 points) on curve and join them, then they will form an closed loop (or we can say convex polygon ) So Centroid of this Polygon , which is surely lies inside the polygon So it's y-coordinate is greater or equal (when all pt's are collinear , I think) to y-coordinates of point on curve which has same x-coordinate as that of centroid of this loop (Polygon) !
Here I used Jenson's inequality
I got 2013 by Cauchy-Scwartz
Log in to reply
Ditto. Do you think solving 4 is enough!Dunno, this waz my only RMO.
I learnt the spelling of "Cauchy Schwartz" today,and I got the solution at my first glance! ;-)
Log in to reply
Please post your method @Ranjana Kasangeri
I think there's no 't'.
Log in to reply
Yeah! You are right. I got the spelling wrong! * sob * @Joel Tan
Sir,Multiply the LHS by summation or (1-x) ,then apply Cauchy!
i got 2013
Log in to reply
Anish can you please post the first 5 questions also... This one I did by cauchy Schwartz
Log in to reply
Its given below the question
Log in to reply
Can someone explain the third question please
Log in to reply
I did it by contradiction but not very sure..
Is that the one about permuting the digits of two powers of 2?
Chebyshev
Log in to reply
Tchebycheff
Log in to reply
-- http://en.wikipedia.org/wiki/Chebyshev%27sinequality
I had given my RMO yesterday and done this using Jensen's inequality.
I used Titu's Lemma for this one!