# How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$

let $x,y,z$ are postive numbers,show that $$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$

My try: Without loss of generality，we assume that $$x+y+z=1$$ and use Holder inequality,we have $$\left(\sum_{cyc}\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}\right)^2\left(\sum_{cyc}((x+y)(x^2+xy+y^2+yz)\right)\ge\left(\sum_{cyc}(x+y)\right)^3$$ then we only prove this $$\dfrac{8}{\sum_{cyc}(x+y)(x^2+xy+y^2+yz)}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$ then I can't

Thank you

• In your approach, you will need to prove $$\dfrac{8}{\sum_{cyc}(x+y)(x^2+xy+y^2+yz)}\ge \left(2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}\right)^2$$, which is more difficult, particularly as it does not hold, for say e.g. $x = \frac5{32},y= \frac7{16}, z= \frac{13}{32}$ – Macavity Feb 13 '14 at 6:34
• x, y and z can be multiplied with a common positive factor without changing the inequality. This allows to assume $x+y+z=1$. But we could also assume $x=1$. – Axel Kemper Feb 13 '14 at 10:56

## 2 Answers

This is a question of the symmetric type, such as listed in:

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

Our function $f$ in this case is: $$f(x,y,z) = \dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}} - \left[2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}\right]$$ Places where $\;|f(x,y,z)|\;$ is less than, say, $\;\epsilon\;$ ("eps", see below), are colored $\color{blue}{blue}$. The blue spots in the pictures are a proof without words that the only minimum is indeed close to zero and near the midpoint $M$ of the equilateral triangle: $(x_M,y_M,z_M) = (1/3,1/3,1/3)$ .

• Picture at top left: geometry of the conditions $\;x,y,z > 0\;$ and $\;x+y+z=1$ .
• Picture at top right: contour lines of $f(x,y,z)$ , as seen in the plane of the $\color{red}{red}$ triangle, are at $25$ equidistant levels, between the minimum and the maximum as found within the viewport
• Picture at bottom left: enlargement of the $\color{blue}{blue\, spot}$ in the picture at top right
• Picture at bottom right: same enlargement but with the value of $\epsilon$ reduced to a much smaller value
A few further details:
• Function values close to viewport minimum are white
• Function values close to viewport maximum are black
• Contour grey values are just the other way around
A = (1,0,0); O = (0,0,0);                        eps := 0.0003;
B = (0,1,0); M = (1/3,1/3,1/3);                  xmin := -0.7; xmax := 0.7;
C = (0,0,1);                                     ymin := -0.6; ymax := 0.8;
xmin := -0.02; xmax := 0.02;                     eps := eps/4096;
ymin := -0.02; ymax := 0.02;

Two or three iterations result in pictures that, quite remarkably, can hardly be distinguished from the two pictures at the bottom:
• Picture at bottom left: enlargement $\times 64$ of the $\color{blue}{blue\,spot}$ in the picture at bottom right
• Picture at bottom right: same enlargement but with the value of $\;\epsilon\;$ reduced to $\;\epsilon/4096\;$
xmin := xmin/64; xmax := xmax/64;                eps := eps/4096;
ymin := ymin/64; ymax := ymax/64;

With calculations ending in a "range check error" because $\epsilon$ ( = eps) has become zero (too) quickly. Which - alright? - is a proof of the statement up to no less than machine precision.

A local and planar orthogonal coordinate system for the equilateral triangle can be obtained as: $$\vec{e}_\xi = \vec{OB}-\vec{OA} = \left[ \begin{array}{c} -1 \\ +1 \\ 0 \end{array} \right] / \sqrt{2}\\ \vec{e}_\eta = \vec{OC}-\left(\vec{OA}+\vec{OB}\right)/2 = \left[ \begin{array}{c} -1/2 \\ -1/2 \\ +1 \end{array} \right] / \sqrt{3/2}$$ Giving for the relationship between global coordinates $(x,y,z)$ and local triangle coordinates $(\xi,\eta)$: $$x = -\frac{1}{\sqrt{2}} \xi -\frac{1}{\sqrt{6}} \eta + \frac{1}{3}\\ y = +\frac{1}{\sqrt{2}} \xi -\frac{1}{\sqrt{6}} \eta + \frac{1}{3}\\ z = \sqrt{\frac{2}{3}} \eta + \frac{1}{3}$$ It's easily verified that $x+y+z=1$ ; the origin of the local coordinate system is at $(\xi,\eta) = (0,0)$ . In this way, the function can be expressed in local triangle coordinates, but (at least for me) that didn't simplify the puzzle much.

A common method to study the function $f(x,y,z)$ in the neighborhood of $M$ is to take derivatives at that point (without a computer algebra system it's a tedious job). To begin with, the zero'th order derivative is zero: $$f(M) = 0$$ The first order derivatives are zero as well, meaning that there is an extremum to be expected there: $$\left[ \begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{array} \right] (M) = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]$$ The second order derivatives are summarized as a Hessian matrix: $$\left[ \begin{array}{ccc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^ f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^ f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^ f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \end{array} \right] (M) = \frac{51}{64} \left[ \begin{array}{ccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array} \right]$$ Therefore $f(x,y,z)(M)$ can be approximated by the following Taylor series expansion: $$f(x,y,z)(M) \approx \frac{51}{32} \left( x^2 + y^2 + z^2 - x y - x z - y z \right)$$ To investigate the nature of this expression, eigenvalues and eigenvectors of the Hessian matrix are determined: $$\left| \begin{array}{ccc} 2-\lambda & -1 & -1 \\ -1 & 2-\lambda & -1 \\ -1 & -1 & 2-\lambda \end{array} \right| = -\lambda^3 + 6 \lambda^2 - 9\lambda = 0 \qquad \Longrightarrow \qquad \lambda \in \left\{ 0,3 \right\}$$ The normed eigenvector corresponding with $\;\lambda=0\;$ is $\;(1,1,1)/\sqrt{3}\;$ and the normed eigenvectors corresponding with $\;\lambda=3\;$ is anything in the plane $\;x+y+z=0\;$ , which is parallel to the plane of our equilateral triangle. So we can choose the local triangle coordinate system for the latter: $$\vec{e}_\xi = \left[ \begin{array}{c} -1 \\ +1 \\ 0 \end{array} \right] / \sqrt{2} \qquad \vec{e}_\eta = \left[ \begin{array}{c} -1/2 \\ -1/2 \\ +1 \end{array} \right] / \sqrt{3/2} \qquad \vec{e}_\zeta = \left[ \begin{array}{c} +1 \\ +1 \\ +1 \end{array} \right] / \sqrt{3}$$ If we express the "conic section" into those local coordinates, then what we get is the following: $$x^2 + y^2 + z^2 - x y - x z - y z = 3 \xi^2 + 3 \eta^2 = \frac{\xi^2}{1/3} + \frac{\eta^2}{1/3} + \frac{\zeta^2}{\infty}$$ The iso-surfaces are a circular cylinder with axis in the direction of $\;\vec{OM} = (1,1,1)/3$ . Thus, as a first approximation, the isolines in the neighborhood of $M$ must be circles, which is exactly what's observed in the above pictures: $$f(x,y,z)(M) \approx \frac{153}{32} \left( \xi^2 + \eta^2 \right) \qquad \mbox{for} \qquad x+y+z=1$$ Behaviour of the abovementioned iterations is explained quite well with this theory. It is also trivial that the only minimum found with this approximation is $\,0\,$ at the midpoint $M$ with $(\xi,\eta) = (0,0)$ .
Guess something more is needed, though, to convert the last argument into a die hard proof.

We can use here the Ji Chen's lemma:

https://artofproblemsolving.com/community/c6h194103 for $$p=\frac{1}{2}.$$

For which it's enough to prove that:

1.$$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq2+\frac{xy+xz+yz}{x^2+y^2+z^2},$$ 2.$$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq1+\frac{2(xy+xz+yz)}{x^2+y^2+z^2},$$ 3. $$\prod_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq\frac{xy+xz+yz}{x^2+y^2+z^2}.$$

A proof of the first inequality.

By C-S $$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq\frac{\left(\sum\limits_{cyc}(x+y)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2+yz)}=\frac{2\sum\limits_{cyc}(x^2+2xy)}{\sum\limits_{cyc}(x^2+xy)}.$$ Now, let $$x^2+y^2+z^2=k(xy+xz+yz).$$

Thus, $$k\geq1$$ and it's enough to prove that $$\frac{2(k+2)}{k+1}\geq2+\frac{1}{k}$$ or $$k\geq1,$$ which is true;

A proof of the second inequality.

By C-S and Rearrangement $$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq\frac{\left(\sum\limits_{cyc}(x+y)(y+z)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}=$$ $$=\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+2x^3y+3x^3z+4x^2y^2+6x^2yz)}=$$ $$=\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)-xyz\sum\limits_{cyc}\left(\frac{x^2}{z}-x\right)}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)}.$$ Id est, it's enough to prove that $$\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)}\geq\frac{(x+y+z)^2}{x^2+y^2+z^2}.$$ Now, let $$x+y+z=3u$$, $$xy+xz+yz=3v^2$$ and $$xyz=w^3$$.

Thus, it's obvious that the last inequality is a linear inequality of $$w^3$$, which says that it's enough to prove the last inequality for an extreme value of $$w^3$$, which happens in the following cases.

a) $$w^3\rightarrow0^+$$.

Let $$z\rightarrow0^+$$ and $$y=1$$.

We need to prove that $$\frac{(x^2+3x+1)^2}{x^4+1+3x^3+3x+4x^2}\geq\frac{(x+1)^2}{x^2+1}$$ or $$x(x^4+x^3-2x^2+x+1)\geq0,$$ which is obvious;

b) Two variables are equal.

Let $$y=z=1$$.

Here we obtain: $$(x-1)^2(2x^3+11x^2+12x+2)\geq0.$$

A proof of the third inequality.

We need to prove that $$(x^2+y^2+z^2)\prod_{cyc}(x+y)^2\geq(xy+xz+yz)\prod_{cyc}(x^2+xy+y^2+yz),$$ which is obviously true after full expanding: $$\sum_{cyc}(x^6y^2+x^5y^3+x^6yz+2x^5y^2z-x^4z^3y-x^4y^2z^2-3x^3y^3z^2)\geq0.$$ Done!

• Yes. This is the first one that Ji Chen's Symmetric Function Theorem applies. I only used it two times. – River Li Jun 29 at 11:00
• I used it in the first time. The respect your of course – Michael Rozenberg Jun 29 at 12:11