[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:
[tex] \frac{bc}{b+c}\leq \frac{b+c}{4}[/tex]
[tex] \frac{ca}{c+a}\leq \frac{c+a}{4}[/tex]
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]
