1) Find the minimum of (a+b)4+(b+c)4+(c+a)4-47(a4+b4+c4) a,b,c are reals.
2) Solve the Diophantine equation for integers x3+2y3+4z3-6xyz=1
Beleive me, number 2 is really hard...
341By minimum I take it you mean a numerical value and also you dont mean a lower bound.
Note that f(0,0,0) = 0
Now if the minimum m = f(p,q,r)
Suppose m>0, then for |k|>1 f(p/k, q/k, r/k) = 1/k4 f(p,q,r) = m/k4<m
Similarly if m<0, we choose |k|>1 and consider f(pk,qk,rk) and exhibit a lower minimum contradicting the fact that m is the minimum value
Hence m = 0
66Those arguments also apply if you replace that factor of 4/7 to something else like 1000....
341That's why I want to know if there's an expression on the RHS like some f(a,b,c)?
66This was one of olympiads postal coaching of last month and the problem statement is exactly what soumik has written.
341Well I have just seen the solution for your 2nd problem in a book. It is, as you said, a tough one. I dont think I should give the source, since the olympiad aspirants are supposed to have a crack at it. But like the 1st problem it seemed like I have seen it somewhere, which seems to be the trend for our olympiad and JEE coaching - the trainers are smug with the solutions in their pockets, ready to be drawn out the moment a student asks them. No real training takes place this way. This is, of course, just my opinion.
341Ah! I have got very good at detective work - i have tracked down the 1st problem too, again from a well known source. So much for original problem setting and solving!
66hmm...so these are copied too... lets hope the students at least are honestly doing them..
341most of them tend to be like this. Even some INMO probs I have seen are from old Hungarian competitions.
