Răspuns:
[tex]\frac{a+b}{c} +\frac{b+c}{a} +\frac{c+a}{a} \geq 6[/tex]
[tex]\frac{a}{c} +\frac{b}{c} +\frac{b}{a} +\frac{c}{a} +\frac{c}{b} +\frac{a}{b} \geq 6[/tex]
[tex](\frac{a}{c} +\frac{c}{a} )+(\frac{b}{c} +\frac{c}{b} )+(\frac{a}{b} +\frac{b}{a} )\geq 6[/tex]
Anaalizam prima paranteza si observam ca-a maai mare decat 2
[tex]\frac{a}{c} +\frac{c}{a} \geq 2[/tex]
duci la acelasi numitor
[tex]\frac{a^2}{ac} +\frac{c^2}{ac} \geq 2\frac{ac}{ac}[/tex]
a²+c²≥2ac
a²-2ac+c²≥0
(a-c)²≥0 evident
Analog demonstrezi c si elelalte paranteze sunt mai mari sau egale cu 2
Deci 2+2+2≥6 Evident
a,b,c>0
Explicație pas cu pas:
