# A Cyclic Inequality from India In Three Variables And More

### Solution

Let's first proof that,

For $a,b\ge 0,$

$\sqrt{a^4+a^2b^2+b^4}\ge a^2+b^2-(2-\sqrt{3})ab,$

with equality at $(k,k),$ $(k,0),$ or $(0,k).$

Indeed, for $x=a^2,y=b^2,$ $\displaystyle s=\frac{1}{2}(x+y),$ $p=\sqrt{xy},$ $s\ge p.$ The inequality reduces to $\sqrt{4s^2-p^2}\ge 2s-(2-\sqrt{3})p$ which is equivalent to $sp\ge p^2.$ If $p=0,$ then equality is obvious. If $p\gt 0,$ $sp\gt p^2$ is implied by $s\ge p.$ Equality in this case is attained for $s=p,$ i.e., $x=y$ which is the same as $a=b.$

The inequality can be rewritten as

$\sqrt{a^4+a^2b^2+b^4}\ge a^2+b^2+4ab-(6-\sqrt{3})ab,$

Now, to obtain the original inequality, we add up

\begin{align} \sqrt{a^4+a^2b^2+b^4}&\ge a^2+b^2+4ab-(6-\sqrt{3})ab\\ \sqrt{b^4+b^2c^2+c^4}&\ge b^2+c^2+4bc-(6-\sqrt{3})bc\\ \sqrt{c^4+c^2a^2+a^4}&\ge c^2+a^2+4ca-(6-\sqrt{3})ca \end{align}

with equality for $(k,k,k),$ $(k,k,0),$ $(k,0,0),$ $k\ge 0,$ and permutations.

### Generalization

The above solution makes it clear that the problem can be extended to a multivariable inequality.

For integer $n\ge 2,$ and non-negative $a_1,a_2,\ldots,a_n,a_{n+1}=a_1,$

$\small{\displaystyle \sum_{cycl}\sqrt{a_k^4+a_k^2a_{k+1}^2+a_{k+1}^4}+(6-\sqrt{3})\sum_{cycl}a_ka_{k+1}\ge 2\left(\sum_{cycl}a_k\right)^2}.$

How would a symmetric extension look like?

### Acknowledgment

The problem has been kindly communicated to me by Leo Giugiuc, along with a solution of his. The problem is by Abhay Chandra (New Delhi). The problem has been originally posted at the Olimpiada pe Scoala (The School Yard Olympiad) facebook group.

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