Let $x_1,x_2,\ldots ,x_{2014}$ be positive real numbers such that
$\large\displaystyle\sum^{2014}_{j=1} x_j=1$
Determine with proof the smallest constant $K$ such that
$\large K \displaystyle\sum^{2014}_{j=1} \dfrac{x^2_j}{1-x_j}\geq1$
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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) !
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$x_{1} + x_{2} + \ldots + x_{2014} = 1$
$\dfrac{x_{1}^{2}}{1 - x_{1}^{2}} + \dfrac{x_{2}^{2}}{1 - x_{2}^{2}} + ........................... + \dfrac{x_{201}4^{2}}{1 - x_{2014}^{2}}$
Applying Titu's Lemma ,
$\dfrac{x_{1}^{2}}{1 - x_{1}^{2}} + \dfrac{x_{2}^{2}}{1 - x_{2}^{2}} + ........................... + \dfrac{x_{201}4^{2}}{1 - x_{2014}^{2}} \geq \dfrac{(x_{1} + x_{2} + .... + x_{2014})^{2}}{2014 - (x_{1} + x_{2} + .....+ x_{2014})}$
$\dfrac{x_{1}^{2}}{1 - x_{1}^{2}} + \dfrac{x_{2}^{2}}{1 - x_{2}^{2}} + ........................... + \dfrac{x_{201}4^{2}}{1 - x_{2014}^{2}} \geq \frac{1}{2013}$
Thus , $\boxed{K = 2013}$
Method 2
$\displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j} - 1}{1 - x_{j}} + \dfrac{1}{1 - x_{j}}$
$\displaystyle \sum_{1}^{2014} -( 1 + x_{j}) + \dfrac{1}{1 - x_{j}}$
$\boxed{- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} = \displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j}}{1 - x_{j}}}$
Applying $A.M \geq H.M$
$\frac{\displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}}}{2014} \geq \dfrac{2014}{ 2014 - ( x_{1} + x_{2} ..... x_{2014})}$
$\displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2}}{ 2013}$
$- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2}}{ 2013} - 2015$
$- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2}- 2015 \times 2013}{2013}$
$- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2} - (2014 + 1)(2014 - 1)}{2013}$
$- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{2014^{2} - (2014^{2} - 1)}{2013}$
$- 2015 + \displaystyle \sum_{1}^{2014} \dfrac{1}{1 - x_{j}} \geq \dfrac{1}{2013}$
$\displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j}}{1 - x_{j}} \geq \dfrac{1}{2013}$
$2013\displaystyle \sum_{1}^{2014} \dfrac{x^{2}_{j}}{1 - x_{j}} \geq 1$
$\boxed{K = 2013}$
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.
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Yes i too enjoyed very much doing by the 2nd method
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Yeah!!
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Its tagged under A.M G.M inequality , have you done using it , if yes then how
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Typo
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u can use \dfrac(it gives bigger fractions) and also \ldots(it gives the dots)
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I applied the \Idots , its not working
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$\ldots$
Its actually small "L"I know its confusing sometimes
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More lines: \rightarrow, \RIghtarrow give us $\rightarrow , \Rightarrow$
Capital Greek alphabet: \gamma, \Gamma give us $\gamma, \Gamma$
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Contrary to @Sandeep Rathod solution there is one more method :
Let an function $y\quad =\quad \cfrac { \quad { x }^{ 2 } }{ 1-x } \\$. in (0,1)
So it inscribe an convex polygon of centroid G so from jensons inequality :
$y\quad =\quad f(x)\quad =\quad \cfrac { \quad { x }^{ 2 } }{ 1-x } \\ \\ { y }_{ G }\quad \ge \quad { y }_{ p }\\ \\ \frac { \sum { f\left( { x }_{ i } \right) } }{ n } \quad \ge \quad f\left( \cfrac { \sum { { x }_{ i } } }{ n } \right) \\ \\ \sum { f\left( { x }_{ i } \right) } \quad \ge \quad nf\left( \cfrac { 1 }{ n } \right) \quad \quad \quad (put\quad n\quad =\quad 2014)\\ \\ 2013\sum { f\left( { x }_{ i } \right) } \quad \ge \quad 1\\ \\ \boxed { K\quad =\quad 2013 }$.
Q.E.D
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i did'nt understood , how it's inscribed? @Deepanshu Gupta
I appreciate - you always think different
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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
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I got 2013 by Cauchy-Scwartz
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Ditto. Do you think solving 4 is enough!Dunno, this waz my only RMO.
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I learnt the spelling of "Cauchy Schwartz" today,and I got the solution at my first glance! ;-)
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Please post your method @Ranjana Kasangeri
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I think there's no 't'.
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Yeah! You are right. I got the spelling wrong! * sob * @Joel Tan
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Sir,Multiply the LHS by summation or (1-x) ,then apply Cauchy!
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i got $2013$
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Anish can you please post the first 5 questions also... This one I did by cauchy Schwartz
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Its given below the question
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Chebyshev
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Tchebycheff
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-- http://en.wikipedia.org/wiki/Chebyshev%27sinequality
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I had given my RMO yesterday and done this using Jensen's inequality.
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I used Titu's Lemma for this one!
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