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Prove that (xyz)1/3[1/x +1/y +1/z] > 3 where x,y, z are +ve real nos.
AM >or= GM
AM of 1/x , 1/y , 1/z = (1/x + 1/y + 1/z) / 3
their GM = (1/xyz)^(1/3)
rearrange to get given expression
GM >= HM
(xyz)1/3 >= 3/(1/x + 1/y + 1/z)
(xyz)1/3 (1/x + 1/y + 1/z) >= 3 .............[proved]
G.P.>H.P.
(xyz)1/3>3/(1/x+1/y+1/z)
Hence proved.
you have mentioned that all are positive real nos...so this gives me a thought about AM greater than or equal to GM.
Consider 1/x, 1/y ,1/z as 3 real nos.... their am is (1/x+1/y+1/z)/ 3 and their GM is cube root of (1/xyz) and since
AM is greater than or equal to Gm
u get their desired result.
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