# A Cyclic But Not Symmetric Inequality in Four Variables

### Problem

### 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.

- A Cyclic But Not Symmetric Inequality in Four Variables
- An Inequality with Constraint
- An Inequality with Constraints II
- An Inequality with Constraint III
- An Inequality with Constraint IV
- An Inequality with Constraint VII
- An Inequality with Constraint VIII
- An Inequality with Constraint IX
- An Inequality with Constraint X
- Problem 11804 from the AMM
- Sladjan Stankovik's Inequality With Constraint
- An Inequality with Constraint XII
- An Inequality with Constraint XIV
- An Inequality with Constraint XVII
- An Inequality with Constraint in Four Variables II
- An Inequality with Constraint in Four Variables III
- An Inequality with Constraint in Four Variables V
- An Inequality with Constraint in Four Variables VI
- A Cyclic Inequality in Three Variables with Constraint
- Dorin Marghidanu's Cyclic Inequality with Constraint
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints II
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints III
- Inequality with Constraint from Dan Sitaru's Math Phenomenon
- Another Problem from the 2016 Danubius Contest
- Gireaux's Theorem
- An Inequality with a Parameter and a Constraint
- Unsolved Problem from Crux Solved
- An Inequality With Six Variables and Constraints
- Cubes Constrained
- Dorin Marghidanu's Inequality with Constraint
- Dan Sitaru's Integral Inequality with Powers of a Function
- Michael Rozenberg's Inequality in Three Variables with Constraints
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints IV
- An Inequality with Arbitrary Roots
- Leo Giugiuc's Inequality with Constraint
- Problem From the 2016 IMO Shortlist
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots II
- A Simplified Version of Leo Giugiuc's Inequality from the AMM
- Kunihiko Chikaya's Inequality $\displaystyle \small{\left(\frac{(a^{10}-b^{10})(b^{10}-c^{10})(c^{10}-a^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge\frac{125}{3}[(a-b)^3+(b-c)^3+(c-a)^3]\right)}$

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