Đáp án:

$c)$ Trước hết, ta chứng minh: $\frac{m}{n}+\frac{n}{m}≥2$ (với $m;n>0)$

Ta có: $\frac{m}{n}+\frac{n}{m}≥2$

$⇔\frac{m^2+n^2}{mn}≥2$

$⇔m^2+n^2≥2mn$

$⇔m^2-2mn+n^2≥0$

$⇔(m-n)^2≥0$ (luôn đúng)

Ta có: $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

$=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1$

$=3+(\frac{a}{b}+\frac{b}{a})+(\frac{b}{c}+\frac{c}{b})+(\frac{a}{c}+\frac{c}{a})$

$≥3+2+2+2=9$ (đpcm)

$b)$ Theo câu a, ta có: $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})≥9$ và $\frac{m}{n}+\frac{n}{m}≥2$

Ta có: $\frac{y}{x}+\frac{x}{y}+\frac{z}{x}+\frac{x}{z}+\frac{y}{z}+\frac{z}{y}≥2+2+2=6$

$⇔(\frac{y}{z}+\frac{x}{z})+(\frac{x}{y}+\frac{z}{y})+(\frac{z}{x}+\frac{y}{x})≥6$

$⇔\frac{y+x}{z}+\frac{x+z}{y}+\frac{z+y}{x}≥6$

Lại có: $(\frac{y}{x+z}+\frac{z}{x+y}+\frac{x}{y+z})(\frac{y+x}{z}+\frac{x+z}{y}+\frac{z+y}{x})≥9$

$⇔(\frac{y}{x+z}+\frac{z}{x+y}+\frac{x}{y+z}).6≥9$ (do $\frac{y}{x+z}+\frac{z}{x+y}+\frac{x}{y+z}>0$)

$⇔\frac{y}{x+z}+\frac{z}{x+y}+\frac{x}{y+z}≥\frac{3}{2}$ (đpcm)