Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.8 AIME, harder Prove it

9・190 Given positive numbers $a, b, c, x, y, z$ and $k$ satisfying
$a+x=b+y=c+z=k$

Prove: $a y+b z+c x<k^{2}$ .

Solution

[Proof] Since $k^{3}=(a+x)(b+y)(c+z)$
$\begin{array}{l} =a b c+a c y+b c x+c x y+a b z+a y z+b x z+x y z \\ =a b c+x y z+k(a y+b z+c x) \end{array}$

Therefore,
$a y+b z+c x<k^{2} \text {. }$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.