RMO/INMO Doubt Board

@Calvin Lin Can you help me with this question. It seems pretty difficult.

Draw the angle bisectors of the lines AB and CD. Let these lines be $L_{1}$ and $L_{2}$.

Then in general, XY will cut each of these lines in one point each. These two points are equidistant from AB and CD as they lie on the angle bisectors.

There will be only one point if XY is parallel to either of $L_{1}$ or $L_{2}$ and thus intersects only one of them.

Clearly, for positive $x$ and $y$,

$4x^{4} + 4y^{3} + 5x^{2} +y + 1 > 6x^2 +6y^2 \geq 12xy$ (By A.M. - G.M. inequality)

I don't think the equality holds though. Kindly check the question again @Shivam Jadhav

First, consider the following Lemma,

Lemma: For positive $x$ and $y$, the following inequality holds,

$xy(x\ln x + y\ln y) \geq \ln x + \ln y$

The equality case being $x=y=1$

Proof: We'll divide the proof into three cases,

Case 1: $x,y \geq 1$

$\implies \ln x,\ln y \geq 0$

$\implies x\ln x \geq \ln x$

$\text{\&}$

$y\ln y \geq \ln y$

$\implies x\ln x + y\ln y \geq \ln x +\ln y$

$\implies xy(x\ln x +y\ln y) \geq \ln x + \ln y$ $( \because x,y \geq 1)$

Similarly, we can proceed with other two cases, i.e.,

Case 2: $x,y < 1$

Case 3:$x<1, y>1$

This proves our Lemma. $\square$

Now, using the Lemma, we have,

$ab(a\ln a + b\ln b) \geq \ln a + \ln b$

$\implies \ln a \times \left(a-\dfrac{1}{ab}\right) + \ln b \times \left(b-\dfrac{1}{ab}\right) \geq 0$

$\implies (a-c) \times \ln a + (b-c) \times \ln b \geq 0$ $(\because abc=1)$

$\implies a\ln a + b\ln b -c \times (\ln a +\ln b) \geq 0$

$\implies a\ln a +b\ln b +c\ln c \geq 0$ $(\because abc=1 \implies \ln a + \ln b + \ln c = 0)$

$\implies \ln (a^a b^b c^c) \geq 0$

$\implies a^a b^b c^c \geq 1$

$\implies a^{b+c} b^{a+c} c^{b+a} \leq 1$

Q.E.D.

Here's a much simpler solution -- We can rewrite the LHS of the inequality as (ac)^b (ab)^c (bc)^a = 1/(a^a b^b c^c) (replacing all the ab's with 1/c's etc.). Now suppose, WLOG, a >= b >= c. Since they multiply to one, a >=1 and c <=1, hence a^a >= a^b (because b <= a), c^c >= c^b, hence a^b b^b c^c >= a^b b^b c^b = 1. Thus 1/(a^a b^b c^c) <= 1.

We can write the following as a summation

$\sum_{k = 0}^{\infty}{\dfrac{n(n+1)(2n+1)}{6 n!}} \\\frac{1}{6}\sum_{k = 0}^{\infty}{\dfrac{2n^3 + 3n^2 + n}{n!}} \\ \frac{1}{3}\sum_{k=0}^{\infty}{\dfrac{n^3}{n!}} + \frac{1}{2}\sum_{k=0}^{\infty}{\dfrac{n^2}{n!}} + \frac{1}{6}\sum_{k=0}^{\infty}{\dfrac{n}{n!}} \\ e \left(\frac{5}{3} + \frac{2}{2} + \frac{1}{6} \right) \\ \boxed{\dfrac{17}{6}e}$

Is the answer $\dfrac{17 \textbf{e}}{6}$

Good question. I show you where I have reached.
$\text{We need to solve this equation to get the value of n} \\ \displaystyle \large n^2 = \omega^{2n-1}$

is there any answer to this?? its very well clear that -1 satisfies the equation but how to find the smallest possible integer, i am clueless

$(1+ \omega)n = \omega^n \\ \text{squaring both sides} \\ (1 + \omega^2 + 2\omega) n^2 = \omega^{2n} \\ \omega n^2 = \omega^{2n} \\ \text{cubing both sides } \\ \omega^3 n^6 = \omega^{6n} \\ \rightarrow n^6 = 1 \\ \text{The smallest solution of this over the reals and satisfying the equation is -1 , so} \\\displaystyle n = \boxed{-1}$

Just curious : How did you get IMO rank 1. I mean do they tell the Rank. I thought they just gave gold medal.
BTW Congratulations for making it to INMO. Any guidance you could give me ?

Try out books like Mathematical Olympiad Challenges by Titu Andreescu or Mathematical Olympiad treasures by the same, Problem solving by Arthur Engel, Rajeev Manocha, Thrills of Pre-College Mathematics( haven't really used it but people say it's good ). Keep practicing RMO papers over and over again and if you have the courage, try out questions from China's Olympiads too.