6
u said one liner so .....lafda kara
did virtually
tellmeh if wrng i'll try:P
edited d post meine x>0 dekha ni tha
find min. value of
[(x+1/x)6-(x6+1/x6)-2]/[(x+1/x)3+(x3+1/x3)]
for x>0
and [.] doesnt mean GINT
6
u said one liner so .....lafda kara
did virtually
tellmeh if wrng i'll try:P
edited d post meine x>0 dekha ni tha
we know the
x+1/x≥1 (AM>GM)
--------
2
x + 1/x ≥2
similarly we get the min value of x6+1/x6≥2
so min value is
[26-2-2]/[8+2]
so the answer is
6
better approach:
put (x+1/x)3=a
and x3+1/x3=b
the expression becomes
a2-b2/a+b
=> a+b
=> 3(x+1/x)
we know that x+1/x≥2
=> expression has min value equal to 6
which is possible when x=1(becoz then only AM=GM)
yeah thats 6 ...............sorry ..........i took x=-1............dekha ni ki x>0
so best approach .put x=1