For all real numbers $x,y,z$ such that $-2\leq x,y,z \leq 2$ and $x^2+y^2+z^2+xyz = 4$, determine the least real number $K$ satisfying \[\dfrac{z(xz+yz+y)}{xy+y^2+z^2+1} \leq K.\]
Problem
Source: Turkey TST 2013 - Day 2 - P3
Tags: inequalities, trigonometry, inequalities proposed
03.04.2013 14:44
An elementary proof:
05.04.2013 18:04
Probably the ugliest inequality I have ever seen at competition.
05.04.2013 18:15
hello, see also here http://www.wolframalpha.com/input/?i=maximize+x*%28x*z%2By*z%2By%29%2F%28x*y%2By%5E2%2Bz%5E2%2B1%29%2Cx%5E2%2By%5E2%2Bz%5E2%2Bx*y*z%3D4%2C-2%3C%3Dx%3C%3D2%2C-2%3C%3Dy%3C%3D2%2C-2%3C%3Dz%3C%3D2 Sonnhard.
05.04.2013 18:24
Well, you can put $x=2\cos A$, $y=2\cos B$, $z=2\cos C$ where $A$,$B$,$C$ are the angles of a triangle, which (when $K=1$) yields \[ 8 \cos A \cos^2 C + 8 \cos B \cos^2 C + 4 \cos B \cos C \le 4 \cos A \cos B + 4 \cos^2 B + 4 \cos^2 C + 1. \] Still sufficiently ugly that I don't want to actually work through it.
06.04.2013 00:47
MathUniverse wrote: Probably the ugliest inequality I have ever seen at competition. In the exam, only two people could predict the answer in this problem. Most of them have thought that the answer was $\frac{3}{4}$. I think this is because of the fact that most inequalities that we see in such exams are symmetric ones having trivial equality conditions and the people should learn how to solve such ugly ones. By the way, I think there are uglier ones in China TSTs.
06.04.2013 01:18
crazyfehmy wrote: MathUniverse wrote: Probably the ugliest inequality I have ever seen at competition. In the exam, only two people could predict the answer in this problem. Most of them have thought that the answer was $\frac{3}{4}$. I think this is because of the fact that most inequalities that we see in such exams are symmetric ones having trivial equality conditions and the people should learn how to solve such ugly ones. By the way, I think there are uglier ones in China TSTs. kardesim sen bunlarla niye tartisiyorsun senin bayragin turkiyenin olmadiise alexandar oyle bir sey soylemezdi hem de esitsizlik guzeldir
06.04.2013 02:23
crazyfehmy wrote: MathUniverse wrote: Probably the ugliest inequality I have ever seen at competition. In the exam, only two people could predict the answer in this problem. Most of them have thought that the answer was $\frac{3}{4}$. I think this is because of the fact that most inequalities that we see in such exams are symmetric ones having trivial equality conditions and the people should learn how to solve such ugly ones. By the way, I think there are uglier ones in China TSTs. ΒΈ I agree that it's nice to see some inequality with a non-standard equality case, but that doesn't mean to create inequality such as expansion of: $(x^5+16x^2-2x^3+2xy^3-5y^9+78y^4-6yz^8-19z^7+2z)^2 \ge 0$ defining that sum of two terms in that expansion equals to a constant.
21.04.2013 19:35
v_Enhance wrote: Well, you can put $x=2\cos A$, $y=2\cos B$, $z=2\cos C$ where $A$,$B$,$C$ are the angles of a triangle, which (when $K=1$) yields \[ 8 \cos A \cos^2 C + 8 \cos B \cos^2 C + 4 \cos B \cos C \le 4 \cos A \cos B + 4 \cos^2 B + 4 \cos^2 C + 1. \] Still sufficiently ugly that I don't want to actually work through it. K=1..When?
22.04.2013 15:26
babystudymath wrote: v_Enhance wrote: Well, you can put $x=2\cos A$, $y=2\cos B$, $z=2\cos C$ where $A$,$B$,$C$ are the angles of a triangle, which (when $K=1$) yields \[ 8 \cos A \cos^2 C + 8 \cos B \cos^2 C + 4 \cos B \cos C \le 4 \cos A \cos B + 4 \cos^2 B + 4 \cos^2 C + 1. \] Still sufficiently ugly that I don't want to actually work through it. K=1..When? See my solution.
22.04.2013 18:00
Crazyfehmy, is there any more natural way to prove this inequality than given sum of squares? I tried Lagrange's multipliers, but it's still very ugly. Thank you in advance.
24.04.2013 00:39
MathUniverse wrote: Crazyfehmy, is there any more natural way to prove this inequality than given sum of squares? I tried Lagrange's multipliers, but it's still very ugly. Thank you in advance. There is another way using the following lemma: Let $x, y, z$ be positive real numbers satisfying $x^2+y^2+z^2+xyz=4$. Then for all real numbers $a, b, c$ we have \[ a^2+b^2+c^2 \geq xab+ybc+zca \] Now, take $a=x+y, \: b=1, \: c=z$. The lemma can be proved using discriminant idea.