Question

Fie a,b,c numere reale pozitive,sa se demonstreze ca :
$\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} \leq \frac{a+b+c}{2}$

Answer 1

[tex] \frac{ab}{a+b} \leq \frac{a+b}{4}\\ 4ab \leq (a+b)^2\\ 4ab\leq a^2+2ab+b^2\\ 0\leq a^2-2ab+b^2\\ (a-b)^2\geq 0(A) [/tex]
Analog se arata ca:
$\frac{bc}{b+c}\leq \frac{b+c}{4}$
$\frac{ca}{c+a}\leq \frac{c+a}{4}$
Le adunam pe toate trei:
[tex]\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\leq \frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4}\\ \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\leq \frac{a+b+b+c+c+a}{4}\\ \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\leq \frac{2a+2b+2c}{4}\\ \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\leq \frac{2(a+b+c)}{4}\\ \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\leq \frac{a+b+c}{2}[/tex]