# Sladjan Stankovik's Inequality With Constraint II

### Problem

### Solution

We need to prove that $3(a^4+b^4+c^4+d^4)+12abcd\ge 24.\,$ By Surányi's Inequality,

$3(a^4+b^4+c^4+d^4)+4abcd\ge (a+b+c+d)(a^3+b^3+c^3+d^3).$

By the Cauchy-Schwarz inequality,

$3(a+b+c+d)(a^3+b^3+c^3+d^3)\ge (a^2+b^2+c^2+d^2)^2.$

Thus, suffice it to show that

$(a^2+b^2+c^2+d^2)^2+8abcd\ge 24.$

Since $a+b+c+d=4,\,$ $a^2+b^2+c^2+d^2=4(1+3t^2),\,$ with $0\le t\le 1.\,$ Hence, we need to prove that

$2(1+3t^2)^2+abcd\ge 3.$

By one of Leonard Giugiuc and Daniel Sitaru's results (Problem 4121, *Crux Mathematicorum*), for $0\le t\le\displaystyle\frac{1}{3},\,$

$abcd\ge (1+t)^3(1-3t)=-3t^4-8t^3-6t^2+1.$

Hence, in this case, suffice it to show that

$2(1+3t^2)^2-3t^4-8t^3-6t^2+1\ge 3.$

But this is equivalent to $t^2(15t^2-8t+6)\ge 0,\,$ which is true.

For $t\ge\displaystyle\frac{1}{3},\,$ $2(1+3t^2)^2+abcd\ge 2(1+3t^2)^2\ge\displaystyle\frac{32}{9}\ge 3.\,$ The proof is complete.

### Acknowledgment

The problem and the above solution have been kindly communicated to me by Leo Giugiuc.

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny