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a fost răspuns

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

Răspuns :

Cpw
[tex](a+b+c)( \frac{1}{a} + \frac{1}{b} + \frac{1}{c})= [/tex]

[tex]=\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}=[/tex]

[tex]=1+ \frac{a}{b} + \frac{a}{c}+ \frac{b}{a} +1+ \frac{b}{c}+ \frac{c}{a} + \frac{c}{b} +1=[/tex]

[tex]=3+ (\frac{a}{b} + \frac{a}{c}+ \frac{b}{a} + \frac{b}{c}+ \frac{c}{a} + \frac{c}{b})=[/tex]

Stim ca media aritmetica ≥ media geometrica
 adica : 

[tex]( \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 [/tex]

=> [tex] (\frac{a}{b} + \frac{a}{c}+ \frac{b}{a} + \frac{b}{c}+ \frac{c}{a} + \frac{c}{b}) \geq 6[/tex]

=> 3+6≥ 9