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Question

Show that
(1) (x+y+z)3>27(y+zx)(z+xy)(x+yz)
(2) xyz>(y+zx)(z+xy)(x+yz).

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Solution

According to AM -GM Inequality,
AMGM
That is arithematic mean is greater than equal to geometric mean.

(y+zx)+(z+xy)+(x+yz)3>((y+zx)(z+xy)(x+yz))13


(x+y+z)3>((y+zx)(z+xy)(x+yz))13


((x+y+z)3)3>((y+zx)(z+xy)(x+yz))


(x+y+z)327>((y+zx)(z+xy)(x+yz))


(x+y+z)3>27(y+zx)(z+xy)(x+yz)




(y+zx)+(z+xy)2>(y+zx)(z+xy)

(y+zx)+(y+xz)2>(y+zx)(y+xz)

(y+xz)+(z+xy)2>(y+xz)(z+xy)

2z2>(y+zx)(z+xy)

z>(y+zx)(z+xy)

y>(y+zx)(y+xz)

x>(y+xz)(z+xy)

Multiply the above three equation, we get

xyz>(y+zx)2(z+xy)2(x+yz)2

xyz>(y+zx)(z+xy)(x+yz)



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