Inegalitate in numerele a,b,c>0

[mihai miculita]

Aratati ca: $\boxed{\frac{a+c}{b+c}+\frac{b+c}{a+c}\le\frac{a}{b}+\frac{b}{a};\,(\forall)\,a,b,c>0}.$

[Ciprian.]

Aratati ca: $\boxed{\frac{a+c}{b+c}+\frac{b+c}{a+c}\le\frac{a}{b}+\frac{b}{a};\,(\forall)\,a,b,c>0}.$

Presupunem $a \ge b=> a^2 \ge b^2, ac \ge bc=> a^2+ac \ge b^2+bc=> \frac{a}{b+c}$$\ge \frac{b}{a+c}=>\frac{a}{b+c} - \frac{b}{a+c} \ge 0(1)$$,a-b \ge 0=> c(a-b) \ge 0 (2)$
Inmultim (1) si (2)$=> (\frac{a}{b+c} - \frac{b}{a+c})c(a-b) \ge 0=> \frac{ac(a-b)}{b+c} + \frac{bc(b-a)}{a+c} \ge 0$$=>\frac{abc+a^2b}{b+c} + \frac{abc+ab^2}{a+c} \le a^2 + b^2$$=>$$\frac{a+c}{b+c}+\frac{b+c}{a+c} \le \frac{a^2+b^2}{ab} = \frac{a}{b} + \frac{b}{a}$ .

[mihai miculita]

Altfel:
$\left(\frac{a}{b}-\frac{a+c}{b+c}\right)+\left(\frac{b}{a}-\frac{b+c}{a+c}\right)=\frac{a(b+c)-b(a+c)}{b(b+c)}+\frac{b(a+c)-a(b+c)}{a(a+c)}=$$\frac{c(b-a)}{b(b+c)}+\frac{c(a-b)}{a(a+c)}$$=c(a-b)\left(\frac{1}{a(a+c)}-\frac{1}{b(b+c)}\right)=$
$=\frac{c(a-b)(ac+a^2-b^2-bc)}{ab(a+c)(b+c)}=\frac{c(a-b)[c(a-b)+(a-b)(a+b)]}{ab(a+c)(b+c)}=$$\frac{c(a-b)^2(a+b+c)}{ab(b+c)(a+c)}\ge 0;(\forall)a,b,c>0.$

[ghenghea1]

Alta...$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \ge \dfrac{c+a}{c+b}+\dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}.$