If x and y are distinct positive numbers, prove that x^-1 y^-1 >(x^2+y^2)^-1?

Answer 1

x^-1 y^-1 > (x^2+ y^2)^-1

=(1/x) (1/y) > 1/(x²+y²)

=1/xy > 1/(x² + y²) .............ans

To check this out,

assuming that x =2, y = 4

1/xy
=1/(2) (4)
=1/8 or 0.125

while,

1/(x²+y²)
=1/(2²+4²)
=1/(4+16)
=1/20 or 0.05

so, it is provenly correct that:

1/xy > 1/(x² + y²)
0.125 > 0.05

Answer 2

The first answer is of course not a proof but merely an illustration that it works with certain numbers.
Since you are told that the numbers are different, replace y
by x + a with a positive. Then multiply out the (x + a)^2 on the right hand side and collect like terms. It is then fairly clear that the denominator of the fraction on the right must be larger than on the left and thus the whole fraction must be smaller. Actually, this proves that the LHS is always more than twice the RHS.

Answer 3

we know
(x-y)^2 >=0
(x^2+y^2 -2xy) > =0

or x^2+y^2 >= 2xy > xy as x and y are positive

taking reciprocal we get

1/(x^2+y^2) < 1/x , 1/y or the given expression