Cyclic Inequality in Four Variables By D. Sitaru

Solution 1

Assume, WLOG, $abcd=1.$ By Bergstrom's inequality,

$\displaystyle \sum_{cycl}\frac{a^7}{a^3+bcd}=\sum_{cycl}\frac{a^8}{a^4+1}\ge\frac{\displaystyle \left(\sum_{cycl}a^4\right)^2}{\displaystyle \sum_{cycl}a^4+4}.$

Setting $\displaystyle x=\sum_{cycl}a^4,$ suffice it to prove that

$x^2-2x-8\ge 0.$

That inequality holds for $x\ge 4$ which is true due to the AM-GM inequality $\displaystyle \sum_{cycl}a^4\ge 4\sqrt[4]{(abcd)^4}=4.$

Equality is attained for $a=b=c=d.$

Solution 2

By the Cauchy-Schwarz inequality,

$\displaystyle LHS=\sum_{cycl}\frac{a^8}{a^4+abcd}\ge\frac{(a^4+b^4+c^4+d^4)^2}{(a^4+b^4+c^4+d^4)+4abcd}.$

Hence, it is enough to show that

$\displaystyle (a^4+b^4+c^4+d^4)^2\ge2abcd(a^4+b^4+c^4+d^4)+8(abcd)^2.$

Indeed, by the AM-GM inequality,

$\displaystyle 2abcd\le\frac{a^4+b^4+c^4+d^4}{2},$

so that

$\displaystyle 8(abcd)^2\le 8\left(\frac{a^4+b^4+c^4+d^4}{4}\right)^2=\frac{(a^4+b^4+c^4+d^4)^4}{2}.$

Equality is attained for $a=b=c=d.$

Acknowledgment

This problem has been kindly posted but Dan Sitaru at the CutTheKnotMath facebook page, with a solution by Le Van (Solution 2). Solution 1 follows in the footsteps of an earlier problem.