Question

1/a+1/b+1/c≥9
Se stie ca a+b+c=1

Answer 1

Explicatie (ca un hint :): Inmultim toata fractia cu (a+b+c)

[tex](a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9 \\ 1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1 \geq 9 \\ (a/b+b/a)+(a/c+c/a)+(b/c+c/b) \geq 6 [/tex] | x (a+b+c)

Inegalitatea de mai sus este adevarata deoarece:
a/b+b/a≥2
a/c+c/a≥2
b/c+c/b≥2 si adunand aceste inegalitati obtinem inegalitatea de mai sus.

Answer 2

Folosim inegalitatea:  x/y + y/x ≥ 2

a + b + c = 1

1/a + 1/b + 1/c  = (a+b+c)/a + (a+b+c)/b + (a + b+c)/c =

= 1+b/a+c/a+a/b+1+c/a +a/c+b/c+1 = 3 +(a/b+b/a)+ (b/c + c/b)+ (c/a+a/c) ≥

≥ 3 + 2  +2 + 2 = 9.