practice problems pending

Use Weighted power mean inequality:

WPM_m = [(w1.a1^m + w2.a2^m..... wn.an^m)/(w1 + w2 ... wn)]^(1/m)

If p > q
then:
WPM_p >= WPM_q

equality holds on a1 = a2 .... an

Q1.
Prove that:
a^4 + b^4 + c^4 >= abc(a+b+c) given a,b,c > 0

S1.
Method 1:
1 step using rearrangement inequality.
Method 2:
[(a^4 + b^4 + c^4)/3]^1/4 >= [(a + b + c)/3]
Take power 4 on both sides:
[(a^4 + b^4 + c^4)/3] >= [(a + b + c)/3]^4 = [(a + b + c)/3]^3.(a+b+c)/3 >= (abc).(a+b+c)/3
=>
(a^4 + b^4 + c^4) >= (abc).(a+b+c) H.P.

Q2. (a, b, c, d, e) are positive real numbers such that a+b+c+d+e=8 and a^2+b^2+c^2+d^2+e^2=16. Find the range of (e).
S2.
a+b+c+d = 8-e
a^2+b^2+c^2+d^2 = 16-e^2
=>
(8-e)/4 <= [(16-e^2)/4]^(1/2)
=>
(8-e)^2/16 = (16-e^2)/4
=>5e^2 - 16e <= 0
=> 0 <= e <= 16/5 = answer

Q3. Find all non-zero real number triplets ((x,y,z)) satisfying 3(x^2+y^2+z^2)=1 and x^2y^2+y^2z^2+z^2x^2 = xyz,(x+y+z)^3.
S3.
Hint 1: prove sigma(x^2.y^2) >= xyz(x+y+z)
Hint 2: (x+y+z)^2 <= 1
Hint 3: xyz(x+y+z) > 0
Hint 4: chain the inequalities and force equality.