practice problems

Q1. (a, b, c) are real numbers such that a+b+c=0, and a^2+b^2+c^2=1. Find the greatest value of a^2*b^2*c^2.
S1.

First a wrong approach:
AM GM on a^2,b^2,c^2
=> (a^2 + b^2 + c^2)/3 >= (abc)^(2/3)
Cube on both sides => (abc)^2 <= 1/27
And equality when a^2 = b^2 = c^2 = 1/3
a+b+c = 0 => WLOG a,b>0 and c<0 => a=b=1/sqrt(3) and c = -1/sqrt(3) but that doesn't satisfy a+b+c = 0 so this this wrong.

WLOG
a,b > 0 and c <0
c^2 = a^2 + b^2 + 2ab
1 = a^2 + b^2 + c^2 = a^2 + b^2 + a^2 + b^2 + 2ab = 2(a^2 + b^2 + ab)
=> a^2 + b^2 + ab = 1/2
AM GM on 
a^2 ,b^2 ,ab => (a^2 + b^2 + ab)/3 >= ab => 1/6 >= ab => ab <= 1/6 and equality on a^2 = b^2 = 1/6
c^2 = a^2 + b^2 + 2ab = 1/2 + ab => c^2 <= 1/2 + 1/6 = 2/3
a^2*b^2*c^2 <= 1/36 * 2/3 = 1/54 = answer.

Method 2:
Also can be solved using vieta's and derivatives.

Q2) (x, y) are positive real numbers. Prove that

4x^4 + 4y^3 + 5x^2 + y + 1 >= 12xy.
S2.
Using AM GM
4x^4 + 1 >= 2.2x^2
4y^3 + y >= 2.2y^2
4x^4 + 4y^3 + 5x^2 + y + 1 >= 4x^2 + 4y^2 + 5x^2 = 9x^2 + 4y^2
Again AM GM on 9x^2, 4y^2 => 9x^2 + 4y^2 >= 2.6.xy = 12xy H.P.

Q3) (x, y, z >= 0). Prove that

x^2 + xy^2 + xyz^2 >= 4xyz - 4.
S3)
Similar to the last question, let's try to reduce powers to get xyz.
x^2 + xy^2 + xyz^2 + 4 >= 4xyz
x^2 + 4 >= 4x
4x + xy^2 >= 4xy
4xy + xyz^2 >= 4xyz
H.P.