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Algebra Difficulty 6.4 National olympiad Prove it

Theorem 1.2 If a,b,ca, b, c are arbitrary real numbers such that a+b+c=pa+b+c=p, then setting ab+bc+ca=p2−q23(q≥0)ab+bc+ca=\frac{p^2-q^2}{3}(q \geq 0) and r=abcr=abc, we have
(p+q)2(p−2q)27≤r≤(p−q)2(p+2q)27\frac{(p+q)^2(p-2q)}{27} \leq r \leq \frac{(p-q)^2(p+2q)}{27}

Solution

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