inequality: sqrt(xy)/z + sqrt(yz)/x + sqrt(xz)/y >= 3
- Thread starter yaszine
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The result follows from the inequality between the arithmetic and geometric mean. Since you must have \(\displaystyle x>0,y>0,z>0\). first multiply both sides by \(\displaystyle xyz\) and divide by \(\displaystyle 3\). Then let \(\displaystyle w_1=(xy)^{3/2},w_2=(yz)^{3/2},w_3=(xz)^{3/2}\). Then note that the left-hand side is simply the arithmetic mean, and the right-hand side is the geometric mean of the \(\displaystyle w_i\).
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Is this part of the arithmetic-geometric mean exercise you're working on in the other thread?
Thank you.
Eliz.
Thank you.
Eliz.