INMO 2009

Q6. Let a, b, c be positive real numbers such that a³ + b³ = c³. Prove that a² + b² - c² > 6(c-a)(c-b)

2nd ones an absolute sitter. a_n = 0 when the number is a perfect square. So a₁a₂... is aperiodic making the number irrational.

i agree but how do u PROVE IT???

bhaiyya i am very poor at graphs.
What is the solution??
I got x = all integers and x belongs to [n, [√(n+1)²+1] -1) for all n belongs to integer and [-1,∞)
when i had solved in the exam

@asish in the 6th one shouldnt it be -c² in the LHS

could be. otherwise its a very weak inequality

Q2. Define a sequence <a_n>_{n=1 to ∞} as follows:
a_n = 0, if the number of positive divisors of n is odd
= 1, if the number of positive divisors of n is even.
(The positive divisors of n include 1 as well as n). Let x=0.a₁a₂a₃... be the real number whose decimal expansion contains a_n in the n-th place, n>=1. Determine, with proof, whether x is rational or irrational.

From rahul's post I gather that we have to prove that if a³+b³ = c³, then

a²+b²-c²> 6(c-a)(c-b)

WLOG, we can assume c=1

So now we have to prove that if a³+b³ = 1, then a²+b²-1 > 6(1-a)(1-b)

Now I substitute as follows p = a+b and q = ab

The inequality now becomes, given

p(p²-3q) = 1, prove that

p²+6p-7>8q

Now since a,b<1, we have a³<a and b³<a. So a+b>a³+b³ = 1. So, we have p>1

From Power Mean Inequality (a³+b³/2)^1/3> (a+b/2).

So p<2^2/3

Now 3q = p - 1/p²

So the inequality becomes to prove that

3p (p+7)(p-1)>8(p³-1)

Since p>1, we can cancel on both sides to reduce it to proving that

3p(p+7)>8(p²+p+1) or 5p²-13p+8<0

This is true if p lies between the roots that is between 1 and 8/5

We have already proved that p>1. Now since p<2^2/3, we have to prove that 2^2/3<8/5

or 5 . 2^2/3<2³

or 5 < 2^7/3 or

125<128 which is obviously true which proves the inequality. This also gives us a very good idea of how tight the given inequality is.

At least one nice problem was there in the INMO!

One point that has to be cleared is how i assumed c = 1

We were given a³+b³ = c³ and asked to prove that

a²+b²-c²>6(c-a)(c-b)

I can write equivalently

(a/c)³+(b/c)³ = 1 and have to prove that (a/c)²+(b/c)²2 - 1 > 6(1-a/c)((1-b/c)

So I can substitute x = a/c and y = b/c and the inequality then is

x³+y³ = 1 and x²+y²-1 > 6(1-x)(1-y)

Instead of all this rigmarole, you can simply state WLOG c = 1 and make life simpler

got to admit that i am actually an ex-iitian. In fact i am hsbhatt from goiit. But i am not any maths graduate. But i just got rehooked on math recently. I like solving probs

u r hsbhatt from goiit!!!!!!! i din know that