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wu :: forums    riddles    medium (Moderators: towr, ThudnBlunder, william wu, Icarus, SMQ, Grimbal, Eigenray)    Prove x^y+y^x>1 « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Prove x^y+y^x>1  (Read 7363 times) cool_joh Newbie     Gender: Posts: 50 Prove x^y+y^x>1   « on: Jun 9th, 2008, 1:48am » Quote Modify Prove that the inequality xy+yx>1 holds for any positive reals x and y. IP Logged MATH PRO Aryabhatta Uberpuzzler     Gender: Posts: 1321 Re: Prove x^y+y^x>1   « Reply #1 on: Jun 9th, 2008, 6:18pm » Quote Modify This below seems to work, but I have a strange feeling something is wrong somewhere...     Assume that both x and y lie in (0,1) (otherwise, we are done).   Now wlog assume that y1/(1-y) >= x1/(1-x)   Note that, this implies that   xy-1/yx-1 >= 1 -- ( A )   Now Consider S =   (xy + yx)1/x   = [yx(1 + xy/yx)]1/x   = y(1 + xy/yx)1/x   Now since 1/x > 1, by Bernoulli's  inequality that (1+x)r >= 1 + rx for x > -1 and r >= 1   we have that     S >= y(1 + xy-1/yx)   = y + xy-1/yx-1 >= y + 1 (by A)   Thus S >= 1 + y > 1   Since S > 1, Sx > 1 and we are done.   « Last Edit: Jun 9th, 2008, 6:20pm by Aryabhatta » IP Logged wonderful Full Member     Posts: 203 Re: Prove x^y+y^x>1   « Reply #2 on: Jun 12th, 2008, 10:31pm » Quote Modify Here is another solution:   Case 1:   Either x,y >= 1 then  x^y >=1  or y^x >=1   => x^y + y^x >1   Case 2: x,y<1   By Bernoulli inequality, we have: x^y = [1 + (x-1)]^y >= 1 + y(x-1)   => x^y + y -1 >=  xy   Likewise: y^x + x-1 >= xy   Thus, if we can show  2xy+1>= x+y the x^y + y^x > 1.   This is true since (x-1)(y-1)+xy>0 given <0x,y<1   Have A Great Day!       « Last Edit: Jun 12th, 2008, 10:54pm by wonderful » IP Logged Aryabhatta Uberpuzzler     Gender: Posts: 1321 Re: Prove x^y+y^x>1   « Reply #3 on: Jun 13th, 2008, 1:12am » Quote Modify on Jun 12th, 2008, 10:31pm, wonderful wrote: Here is another solution:   Case 2: x,y<1   By Bernoulli inequality, we have: x^y = [1 + (x-1)]^y >= 1 + y(x-1)   => x^y + y -1 >=  xy     wonderful, in case the exponent is between 0 and 1, the Bernoulli inequality is reversed, i.e   (1+x)^r <= 1 + rx, so the application here is incorrect.   IP Logged wonderful Full Member     Posts: 203 Re: Prove x^y+y^x>1   « Reply #4 on: Jun 13th, 2008, 1:37pm » Quote Modify Thanks Aryabhatta! You are right.   Have A Great Day! IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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