a ) Ta có : \(S_{\Delta ABC}=S_{\Delta MBC}+S_{\Delta MCA}+S_{\Delta MAB}\)
\(\Rightarrow S_{\Delta ABC}=\frac{1}{2}MD.BC+\frac{1}{2}ME.AC+\frac{1}{2}MF.AB\)
\(\Rightarrow S_{\Delta ABC}=\frac{1}{2}ax+\frac{1}{2}by+\frac{1}{2}cz\)
\(\Rightarrow S_{\Delta ABC}=\frac{1}{2}\left(ax+by+cz\right)\)
\(\Rightarrow2S=ax+by+cz\)
\(\Rightarrowđpcm\)
b ) Ta có :
\(\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\left(ax+by+cz\right)=\left(a^2+b^2+c^2\right)+\left(\frac{a}{x}.by+\frac{b}{y}.ax\right)+\left(by.\frac{c}{z}+cz.\frac{b}{y}\right)+\left(cz.\frac{a}{x}+ax.\frac{c}{z}\right)\)
\(=\left(a^2+b^2+c^2\right)+ab\left(\frac{y}{x}+\frac{x}{y}\right)+bc\left(\frac{y}{z}+\frac{z}{y}\right)+ca\left(\frac{z}{x}+\frac{x}{z}\right)\)
\(\ge a^2+b^2+c^2+2ab+2by+2ca=\left(a+b+c\right)^2\)
( vì ta dễ chứng minh được \(\frac{x}{y}+\frac{y}{x}\ge2\) - tương tự với \(\frac{y}{z}+\frac{z}{y};\frac{z}{x}+\frac{x}{z}\)
Vậy \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\ge\frac{\left(a+b+c\right)^2}{\left(ax+by+cz\right)}=\frac{\left(a+b+c\right)^2}{2S}\)
Dấu " = " xay ra \(\Leftrightarrow x=y=z\)
Vậy Min \(\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=\frac{\left(a+b+c\right)^2}{2S}\)
Chúc bạn học tốt !!
