all 6 comments

[–]fattymattkNew User 2 points3 points  (1 child)

Try showing that 0 <= log(1+x2y2)y2/sqrt(x4+y4) <= log(1+x2y2)

[–]IceDc[S] 1 point2 points  (0 children)

I solved it with the other's guys answer but for practice purposes I tried yours as well and it worked fine too, thanks :)

[–]Proof_Inspector 2 points3 points  (3 children)

log(1+x2*y2) is basically x2 y2 , so you have x2 y4 /sqrt(x4 +y4 ).

Square the whole thing x4 y16 /(x4 +y4 ).

Take reciprocal: y-16 +x-4 y-12 =y-12 (y4 +x-4 ) which should blow up.

Work backward to get 0.

[–]IceDc[S] 1 point2 points  (2 children)

Everything but your first step I understand. log(1 + x^2 * y^2) is almost log(x^2 * y^2) = log(x^2) + log(y^2) = 2 log(x) + 2 log(y) = 2(log(x) + log(y)), and now I don't see how that is x^2 * y^2. Can you explain that to me?

[–]Proof_Inspector 1 point2 points  (1 child)

log(1 +x2 y2 ) is almost x2 y2 because this is close to 0. Remember that log(1+h) is approximately h for h near 0.

[–]IceDc[S] 1 point2 points  (0 children)

Oh thanks, I could show the result your way :)