~ 證不等式 ~

Let a1 = x + 1/x, a2 = y + 1/y and a3 = z + 1/x.

b1 = b2 = b3 = 1

Then by Cauchy-Schwarz's ineq.:

[(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2](12 + 12 + 12) >= (x + y + z + 1/x + 1/y + 1/z)2

3[(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2] >= (1 + 1/x + 1/y + 1/z)2

Now using Arithmetic mean >= Harmonic mean:

(x + y + z)/3 >= 3/(1/x + 1/y + 1/z)

1/3 >= 3/(1/x + 1/y + 1/z)

1/x + 1/y + 1/z >= 9

Hence:

3[(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2] >= (1 + 1/x + 1/y + 1/z)2

= 102

= 100

Finally

(x + 1/x)2 + (y + 1/y)2 + (z + 1/z)2 >= 100/3 = 33 + 1/3

For equality to hold:

In the CS ineq.: x + 1/x = y + 1/y = z + 1/z

In the AM >= HM: x = y = z

Hence equality holds when x = y = z = 1/3

has to sepcify x,y,z are positive.
since x=1;y=1;z=-1 (x+1/x)^2+(y+1/y)^2+(z+1/z)^2 = 12