Problem 75. Suppose that \(a, b, c\) are three positive real numbers verifying
\[
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=13
\]

Find the minimum and maximum values of the expression
\[
P=\frac{a^{3}+b^{3}+c^{3}}{a b c} .
\]
Solution. Denote
$$x=\sum_{c y c} \frac{a}{b} ; y=\sum_{c y c} \frac{b}{a} ; m=\sum_{c y c} \frac{a^{2}}{b c} ; n=\sum_{c y c} \frac{b c}{a^{2}}$$

We have \(x+y=10\) and
$$x^{3}=3(m+n)+6+\sum_{c y c} \frac{a^{3}}{b^{3}} ; y^{3}=3(m+n)+6+\sum_{c y c} \frac{b^{3}}{a^{3}}$$

The identities above yield that
$$x^{3}+y^{3}=\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)+6(m+n)+9$$

We also have
$$m n=3+\sum_{c y c} \frac{a^{3}}{b^{3}}+\sum_{c y c} \frac{b^{3}}{a^{3}}=\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)$$
and
$$x y=\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=3+m+n$$
thus we infer that
$$\begin{array}{c}
10^{3}-30(3+m+n)=m n+6(m+n)+9 \\
\Leftrightarrow 10^{3}-99=m n+36(m+n)
\end{array}$$
so \(m, n\) are two positive roots of the quadratic
$$f(t)=t^{2}-(x y-3) t+(1009-36 x y)$$

Letting now \(r=x y\), we can determine
$$2 \dot{m}=(x y-3) \pm \sqrt{(x y-3)^{2}-4(1009-36 x y)}$$

Consider the function \(g(r)=r-3-\sqrt{r^{2}+138 r-4027}\) for \(0 \leq r \leq 25\). Notice that
$$g^{\prime}(r)=1-\frac{2 r+138}{2 \sqrt{r^{2}+138 r-4027}}<0$$
so we conclude \(11-2 \sqrt{3} \leq m \leq 11+2 \sqrt{3}\), with equality for \(x-y=(a-b)(b-c)(c-a)=0\). The minimum of \(m\) is \(11-2 \sqrt{3}\), attained for \(a=b=(2+\sqrt{3}) c\) up to permutation. The maximum of \(m\) is \(11+2 \sqrt{3}\), attained for \(a=b=(2-\sqrt{3}) c\) up to permutation.