# A Cyclic Inequality in Three Variables XXV

### Solution 1

Let T, A, B, C be points in the plane such that $TA=x;\,$ $TB=y;\,$ $TC=z\,$ and $\angle ATB=\angle BTC=\angle CTA=120^{\circ}.$

$T$ is then the Fermat-Toricelli point of $\Delta ABC.\,$

From the Law of Cosines applied in $\Delta BTC;$ $\Delta CTA;\,$ $\Delta ATB,\,$ respectively,it follows that

\displaystyle \begin{align} a&=BC=\sqrt{y^2+yz+z^2};\\ b&=CA=\sqrt{z^2+zx+x^2};\\ c&=AB=\sqrt{x^2+xy+y^2}. \end{align}

Also, for $S=[\Delta ABC],$ the area of $\Delta ABC,$

$S=\frac{1}{2}TB\cdot TC\sin 120^{\circ}+\frac{1}{2}TC\cdot TA\sin 120^{\circ}+\frac{1}{2}TA\cdot TB\sin 120^{\circ},$

i.e., $\displaystyle S=\frac{\sqrt{3}}{4}(xy+yz+zx).$

From the Hadwiger - Finsler inequality in $\Delta ABC,\,$

\begin{align} &a^2+b^2+c^2\geq 4S\sqrt{3}+(a-b)^2+(b-c)^2+(c-a)^2\;&\Rightarrow\\ &2(ab+bc+ca)\geq 4S\sqrt{3}+a^2+b^2+c^2\;&\Rightarrow\\ &2\sum \sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\\ &\qquad\qquad\geq 4\sqrt{3}\cdot \frac{\sqrt{3}}{4}\sum xy+\sum (x^2+xy+y^2)\;&\Rightarrow\\ &2\sum \sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\\ &\qquad\qquad\geq 3\sum xy+2\sum x^2+\sum xy\;&\Rightarrow\\ &\sum x^2+2\sum xy\leq \sum \sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\;&\Rightarrow\\ &(x+y+z)^2\leq \sum \sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}. \end{align}

The equality holds for $x=y=z.$

### Solution 2

Note that, with $\displaystyle \epsilon=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3},\,$ $\sqrt{x^2+xy+y^2}=|x-y\epsilon|,\,$ $\sqrt{y^2+yz+z^2}=|y-z\epsilon|,\,$ $\sqrt{z^2+zx+x^2}=|z-x\epsilon|,\,$ which reduce the required inequality to

$\displaystyle \sum_{cycl}|(x-y\epsilon)(y-z\epsilon)|\ge (x+y+z)^2.$

Observe that

\begin{align} (x-y\epsilon)(y-z\epsilon)&=xy-zx\epsilon +yz\epsilon^2-y^2\epsilon\\ (y-z\epsilon)(z-x\epsilon)&=yz-xy\epsilon +zx\epsilon^2-z^2\epsilon\\ (z-x\epsilon)(x-y\epsilon)&=zx-yz\epsilon +xy\epsilon^2-x^2\epsilon. \end{align}

Thus,

\begin{align} &\sum_{cycl}|xy-zx\epsilon +yz\epsilon^2-y^2\epsilon|\\ &\qquad \ge |\sum_{cycl}xy-\epsilon\sum_{cycl}xy+\epsilon^2\sum_{cycl}xy-\epsilon\sum_{cycl}x^2|\\ &\qquad=(x+y+z)^2. \end{align}

Equality takes place when $2$ or $3$ of $x,y,z$ is $0,$ or when $x=y=z.$

### Solution 3

Consider the identity

\begin{align} &\sqrt{u^2+uv+v^2)(v^2+vw+w^2)}\\ &\qquad =\sqrt{(v^2+wu+vw)^2+(v^2+uw+vw)(uv-vw)+(uv-vw)^2} \end{align}

that could be verified directly. By setting $x^2+xy+yz=u,\,$ $y^2+xz+yz=v,\,$ $z^2+xy+xz=w;\,$ $xz-xy=a,\,$ $xy-yz=b,\,$ $yz-xz=c\,$ we have $a+b+c=0\,$ and

\displaystyle \begin{align} \sqrt{u^2+ua+a^2} &=\sqrt{\left(u+\frac{a}{2}\right)^2+\frac{3a^2}{4}}\\ &\ge\left|u+\frac{a}{2}\right|\\ &\ge u+\frac{a}{2},\\ \sqrt{v^2+vb+b2} &\ge v+\frac{b}{2},\\ \sqrt{w^2+wc+c2} &\ge w+\frac{c}{2}. \end{align}

From these,

\displaystyle \begin{align} \sum_{cycl}\sqrt{x^2+xy+y^2}\cdot\sqrt{y^2+yz+z^2}&\ge u+v+w+\frac{1}{2}(a+b+c)\\ &=u+v+w=(x+y+z)^2. \end{align}

Equality is attained when two of the three variables are $0$ or when all three are equal.

### Solution 4

\displaystyle \begin{align} &\sum_{cyc}\sqrt{(x^2+xy+y^2)(y^2+yz+z^2)} \\ &\qquad\qquad=\sum_{cyc}\sqrt{\left[\frac{3}{4}(x+y)^2+\frac{1}{4}(y-x)^2\right]\left[\frac{3}{4}(y+z)^2+\frac{1}{4}(y-z)^2\right]} \\ &\qquad\qquad\geq\sum_{cyc}\left[\frac{3}{4}(x+y)(y+z)+\frac{1}{4}(y-x)(y-z)\right]~\text{(Cauchy-Schwarz)}\\ &\qquad\qquad\geq\sum_{cyc}\left(y^2+xz+\frac{y(x+z)}{2}\right) \\ &\qquad\qquad=\left[y^2+z^2+x^2+2(xz+yx+zy)\right]=(x+y+z)^2. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page, with a solution of his (Solution 1) in a LaTeX file. Leo Giugiuc has communicated to me later his solutions (Solution 2 and Solution 3), dated 2012. This was mentioned earlier in a discussion on a similar problem. Solution 4 is by Amit Itagi.