Sagnik just posted this one on my chatbox...
Prove that (a,b, c +ve)
a2+1b+c+b2+1a+c+c2+1a+b ≥3
RMO: 2006
Hint: Nesbitts?
1hey .........
since the expresion symmetric for a,b,c
the minma will occur
wen a=b=c=k(say)
then ur expresion
3/2(k^2+1/k)
differentiate it w.r.t k....
u will get k=1 or -1...........-1 isnt possible as k is +ve........
put k=1........
value of ur expression comes to be 3
62no i did not get it!!
what about (a+b+c)2??
does it attain a minima when a=b=c?
62read post number 4 here http://www.physicsforums.com/showthread.php?t=299490 to get an idea of what i am sayinig.. and why your arguement is not always correct
62I was trying to construct one function but took me long..
try this one..
x2y2+(x-1)2(y-1)2
This is symmetric in x and y...
but is minimum when x=0 and y=1
at x=y, the mimum attained will never be zero
For a 3 variable function, try
f(x,y,z)=x^2y^2z^2+(x-1)^2(y-1)^2(z-1)^2+(x+1)^2(y+1)^2(z+1)^2
This has a minima which takes value 0 at (0,1,-1) or (0, -1, 1) or (1, 0, -1) or (1, -1, 0) or (-1, 0, 1) or (-1, 1, 0) which dont satisfy x=y=z
while the smallest value when x=y=z would be at x=y=z=0 and it would be 2
[1]
62you mean the function that i constructed and the fact that the minimas/ maximas dont necessarily have to be at the points where these functions are all equal?
66One can re-write the LHS as
\dfrac{a\left(a+\frac{1}{a}\right)}{b+c} + \dfrac{b\left(b+\frac{1}{b}\right)}{c+a}+ \dfrac{c\left(c+\frac{1}{c}\right)}{a+b}
Since by AM-GM one has a + 1a ≥ 2 and similarly for the other two, so
\mathrm{LHS}\geq \dfrac{2a}{b+c} + \dfrac{2b}{c+a}+ \dfrac{2c}{a+b}
Now its straight forward Nesbitt's.
