If xyz>0 verify that ((x2)/y)+((z2)/x)+((y2)/z)>x+y+z
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Answer:
x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)
x3+y3+z3-3xyz=0
x3+y3+z3=3xyz
(x3+y3+z3)/xyz=3
(x3/xyz)+(y3/xyz)+(z3/xyz)=3
x2/xy+y2/zx+z2/xy=3
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