(Russia 1195) If x, y > 0, prove that x/(x^4 + y^2) + y/(y^4 + x^2) <= 1/(xy) Thoughts? By the way, this was in Kiran Kedlaya's MOP notes and said to be from Russia 1995, but John Scholes' Kalva archive doesn't have this problem under Russia 1995. Odd. Hint:
HIDE: Click to reveal hidden text This was in the Power Mean Inequality section in the lecture notes.Problem
Source:
Tags: inequalities
Osiris
19.12.2003 02:57
I have a rather unorthodox solution to this:
By substituting a = 1/x and b = 1/y into the inequality, we see that the equality is equivalent to a / (a<sup>4</sup> + b<sup>2</sup>) + b / (b<sup>4</sup> + a<sup>2</sup>) :le: 1/(ab). This is identical to our original inequality. Thus we can assume WLOG that xy = 1. Then,
x / (x<sup>4</sup> + y<sup>2</sup>) + y / (y<sup>4</sup> + x<sup>2</sup>) :le: 1/(xy)
x / (x<sup>4</sup> + 1 / y<sup>2</sup>) + (1 / x) / (1 / x<sup>4</sup> + x<sup>2</sup>) :le: 1
2x<sup>3</sup> / (1 + x<sup>6</sup>) :le: 1
2x<sup>3</sup> :le: 1 + x<sup>6</sup>
This is trivially true.
Nukular
19.12.2003 04:04
Posted: Thu Dec 18, 2003 8:28 pm Post subject: Russia Inequality (Russia 1195) If x, y > 0, prove that x/(x^4 + y^2) + y/(y^4 + x^2) <= 1/(xy)
firstly:
x^4 + y^2 >= 2x^2 * y
and y^ 4 + x ^2 >= 2y^2 * x
so the LHS is <= 1 / (2xy) + 1 / (2xy) = 1 / xy.
I think I've seen this before though... was it from MOP 2002 ? Because I might have seen it in one of the problem sets there...
Fierytycoon
19.12.2003 05:58
Thanks for the solutions. Quote: I think I've seen this before though... was it from MOP 2002 ? Like I said, its in Kedlaya's MOP notes, and so I guess they have been used at MOP before.
belenos
19.12.2003 15:39
Yet an other solution : x^4 + y >= x^3y + xy = xy(x+1) (rearrangement or (x^3-y)(x-y)>= 0) So ineq becomes x/(x+1) + y/(y+1) <= 1 which is true (x+1 >= 2x).
Nukular
20.12.2003 00:36
belenos wrote: Yet an other solution : x^4 + y >= x^3y + xy = xy(x+1) (rearrangement or (x^3-y)(x-y)>= 0) So ineq becomes x/(x+1) + y/(y+1) <= 1 which is true (x+1 >= 2x). Except the first part isnt quite right... take x = 2; y = 3 ; then LHS = 16 + 9 = 25 RHS = 24 + 6 = 30 and LHS < RHS . => <= (Rearrangement only works when x and y (or w/e variables are in the inequality) are symmetric with one another. Since its x^4 + y^2, swapping x and y does NOT give you the same value so its not symmetric. Furthermore, in order to get the contradiction case, you wrote (x^3-y)(x-y) >= 0. Thats also not true, since its easy to get a case where x^3 > y and x < y. (like 2, 3) ) Good try though
belenos
21.12.2003 10:46
Nukular wrote: Except the first part isnt quite right... Oooops :cry: You're right.
Nukular
21.12.2003 17:51
It's alright we've all seen a nice result before and assumed it worked without checking (I definitely have on MULTIPLE occasions) But the thing on this question I like is that we didnt multi-use AMGM too much, since AM-GM tends to make inequalities weaker and weaker the more it is used
Fierytycoon
21.12.2003 18:00
Quote: But the thing on this question I like is that we didnt multi-use AMGM too much, since AM-GM tends to make inequalities weaker and weaker the more it is used How do you mean? Equality is always possible with AM-GM, so I don't understand how it would become weaker.
Arne
21.12.2003 18:17
AMGM is often too strong.
MysticTerminator
21.12.2003 18:26
fyty - just because there is equality at one place does not mean the inequality cannot be made sharper at other places. For example, someone brought up taylor series. If you take just the first term of the taylor series, you have a sorta good approximation to the function. This approximation can be made a lot sharper by taking the first two terms. However, in both approximations, there is still equality at one point (the point at which the taylor series was evaluated). I think this is what's wrong with the "there's still equality" thought. I had trouble with it for a long time and this is the best explanation I could come up with. Hope it's right.
sqing
25.08.2014 04:01
Russia1995: For positive $ x,y$ prove that\[\frac1{xy}\geq \frac x{x^4 + y^2} + \frac y{x^2 + y^4}.\]
AkshajK
25.08.2014 04:06
sqing wrote: Russia1995:For positive $ x,y$ prove that\[\frac1{xy}\geq \frac x{x^4 + y^2} + \frac y{x^2 + y^4}.\]
Expansion gives $x^6 + y^6 + x^2y^2 + x^4y^4 >= x^4y+x^2y^5+x^5y^2+xy^4$. This is simple because \[\dfrac12 x^6 + \dfrac12 x^2y^2 \ge x^4y\]
\[\dfrac12 y^6 + \dfrac12 x^2y^2 \ge xy^4\]
\[\dfrac12 x^6 + \dfrac12 x^4y^4 \ge x^5y^2\]
\[\dfrac12 y^6 + \dfrac12 x^4y^4 \ge x^2y^5\]
Kezer
26.08.2014 15:14
sqing wrote: Russia1995: For positive $ x,y$ prove that\[\frac1{xy}\geq \frac x{x^4 + y^2} + \frac y{x^2 + y^4}.\] Hi, also refer to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=595705