6) Prove that gcd(a,b)×gcd(a,c)×bcd(b,c)≥(gcd(a,b,c)) 3 for all positive integers, a,b , and c . Intuitively, without proof (though if you can provide a proof that would be great), determine the situations where the two sides of this equation are equal. (Note: The originally posed question is relatively easy, so be looking for straight-forward observations about the property of the ged function as opposed to something detailed and esoteric.)

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Let gcd(a,b,c) = m.Therefore, m divides a, b and c.Then, gcd(a,b) = mx, gcd(b,c) = my, gcd(a,c) = mz where x, y, z are all greater than or equal to 1.

Thus, gcd(a,b)xgcd(b,c)x gcd(a,c) = mxmymz = xyzm3

And [gcd(a,b,c)]3 = m3.

xyzm3 is greater than or equal to m3

Therefore, gcd(a,b)xgcd(b,c)x gcd(a,c) is greater than or equal to gcd(a,b,c).

If a/m, b/m, c/m are pairwise coprime, then x=y=z=1.And hence in this case, gcd(a,b)xgcd(b,c)x gcd(a,c) = gcd(a,b,c).