An Application of Schur's Inequality

The following problem has been posted by Leo Giugiuc at the CutTheKnotMath facebook page, with the solution by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc.

For $x,y,z\ge 0,$ the following inequality holds:

$5\left(\sum\sqrt{x+y}\right)\left(\sum\sqrt{(x+y)(y+z)}\right)\\ \ge\left(\sum\sqrt{x+y}\right)^3+18\sqrt{(x+y)(y+z)(z+x)}.$

where all sums are cyclic.

Proof

If $(x+y)(y+z)(z+x)=0,$ then at least two of $x,y,z$ vanish so that the inequality reduces to $10a\sqrt{a}\ge 8a\sqrt{a},$ for some $a\ge 0,$ which is obviously true.

Assume $(x+y)(y+z)(z+x)\ne 0.$ Then $\sqrt{x+y},$ $\sqrt{y+z},$ $\sqrt{z+x}$ form the sides of a triangle. The segments of the sides from the vertices of the triangle to the points of tangency with the incircle split each of the sides into two pats, thus insuring the existence of $a,b,c\gt 0$ such that $\sqrt{x+y}=a+b,$ $\sqrt{y+z}=b+c,$ $\sqrt{z+x}=c+a.$ In terms of $a,b,c$ the inequality becomes

$\begin{align} 5(a+b+c)&[a^2+b^2+c^2+3(ab+bc+ca)]\\ &\ge 4(a+b+c)^3+9(a+b)(b+c)(c+a). \end{align}$

Introduce $S=a^3+b^3+c^3,$ $s=ab(a+b)+bc(b+c)+ca(c+a),$ and $p=abc.$ The straightforward algebraic manipulation reduces the above inequality to

$5S+5s+15s+45p\ge 4S+12s+24p+9s+18p$

which is simplified to $S+3p\ge s.$ The latter is one of the particular cases of Schur's inequality.

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