# Dan Sitaru's Cyclic Inequality In One Variable

### Solution 1

For typographic convenience, let's denote $a=x^2-x+1,\,$ $b=x^2-x+1,\,$ and $c=4x^2+3.\,$ Then, upon squaring the required inequality takes an equivalent form:

$6x^2+8 = 12x^2+10-(6x^2+2) \lt 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}),$

or,

$3x^2+4 \lt \sqrt{ab}+\sqrt{bc}+\sqrt{ca}.$

Squaring once more gives

$(3x^2+4)^2-(ab+bc+ca)\lt 2(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}),$

which reduces to

(*)

$9(x^2+1)\lt 2(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}).$

Now,

\displaystyle\begin{align} a&=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4},\\ b&=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4},\\ c&=4x^2+3\ge 3. \end{align}

The right-hand side of (*) is then estimated as

\displaystyle\begin{align} RHS(*) &\gt 2(a\frac{3}{2}+b\frac{3}{2}+c\frac{3}{4})\\ &=3(2x^2+2)+\frac{3}{2}(4x^2+3)\\ &=12x^2+\frac{21}{2}\\ &\gt 9(x^2+1)=LHS(*). \end{align}

### Solution 2

Let's denote $b=\sqrt{x^2-x+1},\,$ $c=\sqrt{x^2-x+1},\,$ and $a=\sqrt{4x^2+3}.\,$ Note that $a,b,c\,$ are the sides of a triangle: $a+b\gt c,\,$ b+c\gt a,\,c+a\gt b.\,$For example, square$b+c\gt a\,$to obtain$2(x^2+1)+2\sqrt{(x^2+1)^2-x^2}\gt 4x^2+3,\,$or$2\sqrt{(x^2+1)^2-x^2}\gt2x^2+1;\,$squaring the second time:$4x^4+4x^2+4\gt 4x^4+4x^2+1,\,$which is true. In$\Delta ABC,\displaystyle\begin{align} \cos A &= \frac{b^2+c^2-a^2}{2bc}=\frac{-2x^2-1}{2\sqrt{x^4+x^2+1}},\\ \sin A &= \sqrt{1-\left(\frac{-2x^2-1}{2\sqrt{x^4+x^2+1}}\right)^2}\\ &=\sqrt{\frac{3}{4(x^4+x^2+1)}},\\ S &=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot\sqrt{x^4+x^2+1}\cdot\sqrt{\frac{3}{4(x^4+x^2+1)}}\\ &=\frac{\sqrt{3}}{4}. \end{align}$. By the Hadwiger-Finsley inequality,$\displaystyle\sum_{cycl}(a-b)^2+4S\sqrt{3}\lt\sum_{cycl}a^2\,$such that$\displaystyle\sum_{cycl}(a-b)^2+4\sqrt{3}\cdot\frac{\sqrt{3}}{4}\lt x^2-x+1+x^2+x+1+4x^2+3,$i.e.,$\displaystyle\sum_{cycl}(a-b)^2\lt 6x^2+2.\,\$ This is exactly the required inequality.

### Acknowledgment

Dan Sitaru has kindly posted a problem from the Romanian Mathematical Magazine, with a solution (Solution 1) by Soumava Chakraborty. Solution 2 is by Dan Sitaru; Solution 3 is by Nassim N. Taleb.