A Cyclic Inequality from India In Three Variables And More

Problem

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.

 

Cyclic inequalities in three variables

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