Problem 1120

13.12. 1) By the mean value theorem, we have

$(a b+b c+c a)(a+b+c)=a^{2} b+b^{2} a+c^{2} a+a^{2} c+b^{2} c+c^{2} b+3 a b c \geqslant$

$\geqslant 6 a b c+3 a b c=9 a b c$,

from which the inequality follows

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{9}{P}$$

2) The inequality

$$a^{2}+b^{2}+c^{2} \geqslant P^{2} / 3$$

follows from the chain of relations

$$\begin{aligned} & P^{2}=(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c \leqslant \\ & <a^{2}+b^{2}+c^{2}+\left(a^{2}+b^{2}\right)+\left(a^{2}+c^{2}\right)+\left(b^{2}+c^{2}\right)=3\left(a^{2}+b^{2}+c^{2}\right) . \end{aligned}$$

3) By Heron's formula and the mean value theorem, we get

$$\begin{aligned} 27 S^{2}=27(P / 2) & (P / 2-a)(P / 2-b)(P / 2-c) \leqslant \\ & \quad \leqslant 27(P / 2)\left(\frac{(P / 2-a)+(P / 2-b)+(P / 2-c)}{3}\right)^{3}=P^{4} / 16 \end{aligned}$$

i.e., $P^{2} \geqslant 12 \sqrt{3} S$.

4) According to inequalities 2) and 3), we have

$$a^{2}+b^{2}+c^{2} \geqslant P^{2} / 3 \geqslant 4 \sqrt{3} S$$

5) The inequality

$$a^{3}+b^{3}+c^{3} \geqslant P^{3} / 9$$

follows from the chain of relations

$P^{3}=(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+3 a b(a+b)+3 b c(b+c)+$ $+3 c a(c+a) \leqslant a^{3}+b^{3}+c^{3}+2\left(a^{3}+b^{3}+c^{3}\right)+3\left(a^{2}-a b+b^{2}\right)(a+b)+$ $+3\left(a^{2}-a c+c^{2}\right)(a+c)+3\left(b^{2}-b c+c^{2}\right)(b+c)=9\left(a^{3}+b^{3}+c^{3}\right)$.

6) According to inequalities 3) and 5), we have

$$a^{3}+b^{3}+c^{3} \geqslant P^{3} / 9 \geqslant 12 \sqrt{3} S P / 9=(4 \sqrt{3} / 3) S P$$

7) From inequality 4), we get

$$\begin{aligned} & 16 S^{2} \leqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{3}= \\ & =\frac{a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}}{3} \leqslant a^{4}+b^{4}+c^{4} \end{aligned}$$