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If a, b, x, y are positive natural numbe...

If a, b, x, y are positive natural numbers such that `(1)/(x) + (1)/(y) = 1` then prove that `(a^(x))/(x) + (b^(y))/(y) ge ab`.

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Consider the positive number `a^(x), a^(x),....y` times and `b^(y), b^(y),....x` times
For all these numbers,
`AM = ({a^(x) + a^(x) + ...y " times"} + {b^(y) +b^(y) + .... x " times"})/(x + y) = (ya^(x) + xa^(y))/((x + y))`
`GM = {(a^(x). a^(x)...y " times")(b^(y).b^(y)....x " times")}^((1)/((x + y))) = [(a^(xy)).(b^(xy))]^((1)/((x + y))) = (ab)^((xy)/((x + y)))`
As `(1)/(x) + (1)/(y) = 1, (x + y)/(xy) = 1`, i.e., `x + y = xy`
So using `AM ge GM (ya^(x) + xa^(y))/(x + y) ge (ab)^((xy)/(x + y))`
`:. (ya^(x) + xa^(y))/(xy) ge ab " or " (a^(x))/(x) + (a^(y))/(y) ge ab`
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