\hspace{-16}$If $\mathbf{a+b+c=1}$ and $\mathbf{a^2+b^2+c^2=2}\;$\\\\ Then Range of $\mathbf{a^4+b^4+c^4}$
1HINT:
a4+b4+c4=p
{a2+b2}2-2(a2b2)=p-c4
(2-c2)2-2[(1-c2)-(a2+b2)2]=p-c4
Then p=[(4-4c2+c4)-(1-c2)-(2-c2)+c4]
Now the expr. is a function of c...and can be further factorised....
Hope there are no silly mistakes.... if so....please forgive me.....
Sorry in advance
1There has been a typing mistake....
The expr. should have been...
(2-c2)2-2{(1-c)2-(2-c2)2}2=p-c4
Sorry for the inconvenience.....
3The square of a^2+b^2+c^2 is a^4+b^4+c^4+2(a^2*b^2+a^2*c^2+b^2*c^2)=4, therefore
a^4+b^4+c^4=4-2(a^2*b^2+a^2*c^2+b^2*c^2)
Using AM-GM, (a^2*b^2+a^2*c^2+b^2*c^2)/3 is >or=to (the cube root of a^2*b^2*a^2*c^2*b^2*c^2),
a^2*b^2+a^2*c^2+b^2*c^2 is >or=to 3*(the cube root of a^4*b^4*c^4), so the maximum value is
4-6*(the cube root of a^4*b^4*c^4)
using the AM-GM on a^4+b^4+c^4, you get a^4+b^4+c^4>or=to 3*(the cube root of a^4*b^4*c^4), the minimum value.
341We have using the given conditions
a^4+b^4+c^4 = 4abc + \frac{7}{2}
So we need to find the range of abc
Now we also get ab+bc+ca = -\frac{1}{2}
So a,b,c are roots of x^3-x^2-\frac{x}{2}+t=0
and t =-abc
Now the numbers that come are not nice, so i will give an outline:
1. find the max and min of the polynomial
2. impose the condition that max>0 and min <0 to get bounds for t and hence for abc
3. now find the max and min for given expression
I did a calculation and obtained (2.19, 2.93)
hope i am right :D
