# A Problem form the Short List of the 2018 JBMO

### Solution

The required inequality is equivalent to

$(ab^3+bc^3+cd^3+da^3)^2\ge (a^2b^2+b^2c^2+c^2d^2+d^2a^2)^2.$

By the Cauchy-Schwarz inequality,

$(ab^3+bc^3+cd^3+da^3)(a^3b+b^3c+c^3d+d^3a)\ge (a^2b^2+b^2c^2+c^2d^2+d^2a^2)^2.$

Hence suffice it to prove that

$(ab^3+bc^3+cd^3+da^3)^2\ge (ab^3+bc^3+cd^3+da^3)(a^3b+b^3c+c^3d+d^3a)$

i.e.,

$ab^3+bc^3+cd^3+da^3\ge a^3b+b^3c+c^3d+d^3a.$

Rewrite the inequality successively:

\begin{align} &ab^3+bc^3+cd^3+da^3\ge a^3b_b^3c+c^3d+d^3a,\\ &a(b^3-d^3)+b(c^3-a^3)+c(d^3-b^3)+d(a^3-c^3)\ge 0,\\ &(a-c)(b^3-d^3)-(b-d)(a^3-c^3)\ge 0,\\ &(a-c)(b-d)(b^2+bd+d^2-a^2-ac-c^2)\ge 0. \end{align}

The last inequality is true because $(a-c)(b-d)\ge 0$ and $(b^2-a^2)+(bd-ac)+(d^2-c^2)\ge 0.$ Equality is attained when $a=c,$ or $b=d,$ or $(a=b)\text{ and }(c=d).$ Combining these with the equality cases in the Cauchy-Schwarz inequality, we that in the original inequality, equality is attained iff $a=b=c=d.$

Remark

We could have proved $ab^3+bc^3+cd^3+da^3\ge a^3b_b^3c+c^3d+d^3a$ differently:

\begin{align} &2(ab^3+bc^3+cd^3+da^3)\ge ((ab^3+bc^3+cd^3+da^3))+(a^3b_b^3c+c^3d+d^3a)\\ &=(ab^3+a^3b)+(bc^3+b^3c)+(cd^3+c^3d)+(da^3+d^3a)\\ &ge 2a^2b^2+2b^2c^2+2c^2d^2+2d^2a^2, \end{align}

by the AM-GM inequality, with the same conclusion.

### Acknowledgment

The problem and the above solution are by Leonard Giugiuc and Andrei Eckstein, Romania

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Copyright © 1996-2018 Alexander Bogomolny
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