# Lorian Saceanu's Cyclic Inequalities with Three Variables

### Solution, Question 1

Let $a=x+y,$ $b=x+z,$ $c=y+z.$ From the premises, $x,y,z\gt 0.$ Substituting and carrying out algebraic operations, the problem reduces to

$x^4z+z^4y+y^4x\ge xyz(x+y+z).$

This in turn is equivalent to

$\displaystyle \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x}\ge xy+yz+zx.$

Now, due to Bergstrom's inequality,

$\displaystyle \sum_{cycl}\frac{x^3}{y}\ge\frac{(x^2+y^2+z^2)^2}{xy+yz+zx}.$

Thus, suffice it to prove that

$\displaystyle \frac{(x^2+y^2+z^2)^2}{xy+yz+zx}\ge xy+yz+zx.$

But this is obvious because $x^2+y^2+z^2\ge xy+yz+zx.$

### Solution, Question 2

It can be verified that

\displaystyle\begin{align} &2+\sum_{cycl}\frac{a}{b}-\sum_{cycl}\frac{a}{c}-2\frac{ab+bc+ca}{a^2+b^2+c^2}\\ &\qquad\qquad=\frac{1}{a^2+b^2+c^2}\sum_{cycl}\frac{(c+a-b)(a+b-c)(a-b)^2}{ab}, \end{align}

which shows that the two questions are equivalent.

### Acknowledgment

The two problem (with solutions) have been kindly communicated to me by their author Lorian Saceanu.

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