# A Cyclic Inequality in Three Variables XXIII

### Solution 1

\begin{align} &(ab-bc)^\ge 0\,\Longleftrightarrow\\ &a^2b^2+b^2c^2\ge 2ab^2c.\, \text{Similarly,}\\ &b^2c^2+c^2a^2\ge 2abc^2\,\text{and}\\ &c^2a^2+a^2b^2\ge 2a^2bc. \end{align}

Adding up, $a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c).\,$ It's not hard to see that this inequality is equivalent to the original one (just carry out the implied arithmetic and cancel similar terms.)

### Solution 2

We'll start with $a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c).\,$ This is equivalent to

$\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c.$

Assume, WLOG, that $a\ge b\ge c.\,$ Then $ab\ge ac\ge bc\,$ and $\displaystyle \frac{1}{c}\ge\frac{1}{b}\ge\frac{1}{a}.\,$ By the Rearrangement inequality, it then follows that

\displaystyle \begin{align} \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&=(ab)\cdot\frac{1}{c}+(bc)\cdot\frac{1}{a}+(ac)\cdot\frac{1}{b}\\ &=(ab)\cdot\frac{1}{c}+(ac)\cdot\frac{1}{b}+(bc)\cdot\frac{1}{a}\\ &\ge (ab)\cdot\frac{1}{b}+(ac)\cdot\frac{1}{a}+(bc)\cdot\frac{1}{c}\\ &= a+b+c. \end{align}

### Solution 3

In simplification of the above argument, observe that $x^2+y^2+z^2\ge xy+yz+zx\,$ and let $x=bc\,$ $y=ca\,$ $z=ab\,$ to obtain $a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c).\,$

### Solution 4

Another way of looking at $a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c):\,$ This is true due to Muirhead's theorem because the triple $(2,2,0)\,$ majorizes $(2,1,1).$

### Acknowledgment

Dan Sitaru has kindly posted the problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later added the solution (Solution 1) by Abdul Aziz. Solution 3 has been posted independently by Kevin Soto Palacios and Soumava Chakraborty; Solution 4 is by Amit Itagi. Kevin Soto Palacios has observed that the constraint $a,b,c\gt 0\,$ is unnecessary.