Bài giải

Ta có : ${{S}_{1}}+{{S}_{2}}+{{S}_{3}}$

$=\left( \frac{b}{a}.x+\frac{c}{a}.z \right)+\left( \frac{a}{b}.x+\frac{c}{b}.z \right)+\left( \frac{a}{c}.z+\frac{b}{c}.y \right)$

$=\left( \frac{b}{a}.x+\frac{a}{b}.x \right)+\left( \frac{b}{c}.y+\frac{c}{b}.y \right)+\left( \frac{a}{c}.z+\frac{c}{a}.z \right)$

$=\left( \frac{a}{b}+\frac{b}{a} \right).x+\left( \frac{b}{c}+\frac{c}{b} \right).y+\left( \frac{a}{c}+\frac{c}{a} \right).z$

Áp dụng bất đẳng thức Cô-si ta có :

$\frac{a}{b}+\frac{b}{a}\ge 2\sqrt{\frac{a}{b}.\frac{b}{a}}=2$

$\frac{b}{c}+\frac{c}{b}\ge 2\sqrt{\frac{b}{c}.\frac{c}{b}}=2$

$\frac{a}{c}+\frac{c}{a}\ge 2\sqrt{\frac{a}{c}.\frac{c}{a}}=2$

Suy ra ${{S}_{1}}+{{S}_{2}}+{{S}_{3}}\ge 2x+2y+2z=2\left( x+y+z \right)=2.5=10$

Vậy ${{S}_{1}}+{{S}_{2}}+{{S}_{3}}\ge 10$

Hay

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