# An Inequality with Constraint in Four Variables IV

### Solution 1

$(x+y)^3=x^3+y^3+3xy(x+y).\,$ With $x=a+b\,$ and $y=c+d\,$ we have

\displaystyle \begin{align} &\left(\sum_{cycl}a\right)^3=(a+b)^3+(c+d)^3+3(a+b)(c+d)\sum_{cycl}a&\Leftrightarrow\\ &27=\sum_{cycl}a^3+3ab(a+b)+3cd(c+d)+9(a+b)(c+d)&\Leftrightarrow\\ &27=\sum_{cycl}a^3+3ab(a+b)+3cd(c+d)+9(ac+ad+bc+bd)&\Leftrightarrow\\ &27+3\sum_{cycl}abc=\sum_{cycl}a^3+3ab\sum_{cycl}a+3cd\sum_{cycl}a+9(ac+ad+bc+bd)&\Leftrightarrow\\ &27+3\sum_{cycl}abc=\sum_{cycl}a^3+9\sum_{all}ab. \end{align}

Suffice it to show that $\displaystyle 9\sum_{all}ab\ge 54\sqrt{abcd}.\,$ But this is true by the AM-GM inequality.

### Solution 2

Denote $\displaystyle s_3=\sum_{cycl}abc\,$ and $\displaystyle S_3=\sum_{cycl}a^3\,$ and homogenize using the constraint:

$\displaystyle \left(\sum_{cycl}a\right)^3+3s_3\ge S_3+18\left(\sum_{cycl}a\right)\sqrt{abcd}.$

Now, recollect one of Newton's identities:

$\displaystyle S_3=\left(\sum_{cycl}a\right)\left(\sum_{cycl}a^2\right)-\left(\sum_{cycl}a\right)\left(\sum_{all}ab\right)+3s_3.$

This leadds to an equivalent inequality:

$\displaystyle \small{\left(\sum_{cycl}a\right)^3+\left(\sum_{cycl}a\right)\left(\sum_{all}ab\right)-\left(\sum_{cycl}a\right)\left(\sum_{cycl}a^2\right)\ge 18\left(\sum_{cycl}a\right)\sqrt{abcd}}.$

This is equivalent to

$\displaystyle \left(\sum_{cycl}a\right)\left[3\left(\sum_{all}ab\right)\right]\ge 18\left(\sum_{cycl}a\right)\sqrt{abcd}$

and, finally, to

$\displaystyle \sum_{all}ab\ge 6\sqrt{abcd}$

which is true by the AM-GM inequality.

Note that equality is reached for $\displaystyle a=b=c=d=\frac{3}{4}\,$ or $(3,0,0,0)\,$ and permutations.

### Acknowledgment

This problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Leo Giugiuc.