3rd USAMO 1974

Show that for any positive reals x, y, z we have x^xy^yz^z ≥ (xyz)^a, where a is the arithmetic mean of x, y, z.

Solution

Without loss of generality x ≥ y ≥ z. We have x^xy^y ≥ x^yy^x, because that is equivalent to (x/y)^x ≥ (x/y)^y which is obviously true. Similarly y^yz^z ≥ y^zz^y and z^zx^x ≥ z^xx^z. Multiplying these three together we get (x^xy^yz^z)^x ≥ x^y+zy^z+xz^x+y. Multiplying both sides by x^xy^yz^z gives (x^xy^yz^z)³ ≥ (xyz)^3a. Taking cube roots gives the required result.

3rd USAMO 1974

© John Scholes
jscholes@kalva.demon.co.uk
19 Aug 2002