Question

Daca a,b,c sunt numere reale pozitive, sa se arate ca (a+b+c)(1/a+1/b+1/c)>9

Answer 1

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

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

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

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

Stim ca media aritmetica ≥ media geometrica
 adica : 

$( \frac{a}{b} + \frac{a}{c}+ \frac{b}{a} + \frac{b}{c}+ \frac{c}{a} + \frac{c}{b} ) * \frac{1}{6} \geq \sqrt[6]{ \frac{a}{b} * \frac{a}{c}* \frac{b}{a} * \frac{b}{c}* \frac{c}{a} * \frac{c}{b} }=1$

=> $(\frac{a}{b} + \frac{a}{c}+ \frac{b}{a} + \frac{b}{c}+ \frac{c}{a} + \frac{c}{b}) \geq 6$

=> 3+6≥ 9