Inequality

a)Since 2(x^2+y^2)-(x+y)^2 = (x-y)^2 >=0

therefore Å2(x^2+y^2) >= x+y -----(1)

and since (x+y)^2 >= (x+y)^2 -(x-y)^2 = 4xy

therefore x+y >=2Åxy ----(2)

from (1) and (2), we have

Å2(x^2+y^2) >= x+y>= 2Åxy


b)
i) since Å2(x^2+y^2) >= x+y
put x=a and y=b, we have Å2(a^2+b^2) >= a+b -----(3)
put x=b and y=c, we have Å2(b^2+c^2) >= b+c -----(4)
put x=c and y=a, we have Å2(c^2+a^2) >= c+a ------(5)

(3)+(4)+(5), we have

Å2(a^2+b^2) +Å2(b^2+c^2) +Å2(c^2+a^2) >= 2(a+b+c)

dividing both sides by Å2

Å(a^2+b^2) +Å(b^2+c^2) +Å(c^2+a^2) >=Å2(a+b+c)

ii) since x+y>= 2Åxy

put x=a, y=b, we have a+b>= 2Åab -----(6)
put x=b, y=c, we have b+c>= 2Åbc -----(7)
put x=c, y=a , we have c+a>= 2Åca -----(8)

(6)*(7)*(8), we have

(a+b)(b+c)(c+a)=8abc

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