# A Cyclic But Not Symmetric Inequality in Four Variables

### Solution

\begin{align} &5(a+b+c+d)+\frac{26}{abc+bcd+cda+dab}\\ &\qquad \ge 5(a+b+c+d)+\frac{26}{\displaystyle \frac{(a+b)^2}{4}(c+d)+\frac{(c+d)^2}{4}(a+b)}\\ &\qquad =5(a+b+c+d)+\frac{104}{(a+b)(c+d)(a+b+c+d)}\\ &\qquad \ge 5(a+b+c+d)+\frac{104}{\displaystyle \frac{(a+b+c+d)^2}{4}(a+b+c+d)}\\ &\qquad =5(a+b+c+d)+\frac{416}{(a+b+c+d)^3}. \end{align}

Let $a+b+c+d=t.\,$ $t\ge 4,\,$ because of the AM-GM inequality and the condition $abcd=1.\,$

We have to prove that $\displaystyle 5t+\frac{416}{t^3}\ge 26.5,\,$ provided $t\ge 4.\,$ We have a sequence of equivalences:

\displaystyle \begin{align} &5t+\frac{416}{t^3}\ge 26.5\,\Longleftrightarrow\\ &\qquad 10t^4-53t^3+832\ge 0\,\Longleftrightarrow\\ &\qquad (t-4)(10t^3-13t^2-52t-208)\ge 0\,\Longleftrightarrow\\ &\qquad (t-4)[10t^2(t-4)+27t(t-4)+56(t-4)+16]\ge 0, \end{align}

which is true. Equality is attained for $a=b=c=d=1.$

### Acknowledgment

The link to the problem above has been posted on the CutTheKnotMath facebook page by Leo Giugiuc. The problem has been devised by Richdad Phuc following the ideas of Leo Giugiuc. The solution is by Ghimisi Dumitrel.