Integer and common expression

Let $A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ . We can simplify it to $A=3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}$ .

Applying the Cauchy inequality multiple times for $a, b, c > 0$ : $\frac{a}{b}+\frac{b}{a}\ge 2\sqrt{\frac{a}{b}\times \frac{b}{a}}=2$ , so $\frac{b}{c}+\frac{c}{b}\ge 2$ and $\frac{c}{a}+\frac{a}{c}\ge 2$ .

Because $A=3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}$ , the smallest value of $A$ is $3+2+2+2=9$ , or $A\ge 9$ .

Next, as we have to find the integer part of $A$ and we have $A\ge 9$ , we will have to prove that $A<10$ too.

Assume that $1<a\le b\le c<2$ (all other cases of the order of $a, b, c$ have similar proofs), in this case, $a-b\le 0$ and $b-c\le 0$ . This leads to $\left(a-b\right)\left(b-c\right)\ge 0$ , or $ab-ac-b^2+bc\ge 0$ , implying that $ab+bc\ge b^2+ac$ .

Divide the inequality above by $ab$ and $bc$ , we have this set of two inequalities: ${\begin{cases}1+\frac{c}{a}\ge \frac{b}{a}+\frac{c}{b}\\1+\frac{a}{c}\ge \frac{b}{c}+\frac{a}{b}\end{cases}}$

Add both inequalities together, we will have $\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}\le \frac{a}{c}+\frac{c}{a}+2$ .

Because $A=3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}$ , $A\le 3+\frac{a}{c}+\frac{c}{a}+2+\frac{a}{c}+\frac{c}{a}$ or $A\le 5+2\left(\frac{a}{c}+\frac{c}{a}\right)$ .

At this point, we must remember that $1<a\le b\le c<2$ , because $a,b,c>0$ we have: $c \ge a$ so $\frac{c}{a}>1>\frac{1}{2}$ , we also have $\frac{c}{a}<\frac{2}{a}<2$ because $c<2$ and $a>1$ . This implies that $\frac{c}{a}-\frac{1}{2}>0$ and $\frac{c}{a}-2<0$ , which leads to $\left(\frac{c}{a}-\frac{1}{2}\right)\left(\frac{c}{a}-2\right)<0$ or $\left(\frac{c}{a}\right)^2-\frac{5}{2}\times \frac{c}{a}+1<0$ .

Divide the entire inequality above by $\frac{c}{a}$ (we are able to do this because $a,b,c\ne 0$ ), we will have $\frac{c}{a}-\frac{5}{2}+\frac{a}{c}<0$ or $\frac{c}{a}+\frac{a}{c}<\frac{5}{2}$ . Because $A\le 5+2\left(\frac{a}{c}+\frac{c}{a}\right)$ as proven above, $A<5+2\times \frac{5}{2}$ or $A<10$ .

Combine with $A\ge 9$ , we will have $9\le A<10$ .

So $P=\left[A\right]=9$ .