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| If x and y are positive |
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Captain
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I dont know a fastest way. But this is how I thought - First I note that 1/root(x+y) is symmetric wrt x,y
This is not symmetric with respect to x,y. So this cant be correct. What if I take a very small value of y; 1/root(x+y) goes to 1/root(x). but this equation goes to 1/(2root(x)). if x = 4 then 1/root(x+y) = 1/2, but I. becomes 1/4. Thus not always true.
This one is symmetric. so I will use x = y; III. becomes 0, and 1/root(x+y) becomes 1/root(2x); III will be less atleast in one case. Now
This one is also symmetric. And putting x=y we will get II = 1/root(x) and the other as 1/ {root(2)*root(x)}, Thus II is greater. At this point I declared II as the answer. To be sure I wrote both the expressions and proved II>1/root(x+y), for generic cases. |
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Ron Purewal
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