An elementary proof:

Probably the ugliest inequality I have ever seen at competition.

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.

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.

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.

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

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. ¸ 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.

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?

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.

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.

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.