Question

Fie numerele a,b,c,numere reale pozitive cu a·b·c=1.Demonstrati inegalitatea
$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \leq \frac{ a^{2} + b^{2}+ c^{2} }{2}$

Answer 1

vom arata ca $\frac{xy}{x+y} \leq \frac{x+y}{4}$, pentru orice x,y numere reale pozitive

relatia este echivalenta cu (x+y)² ≥ 4xy  ⇔ (x-y)² ≥ 0 care este adevarata.acum

$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} = \frac{c}{ac+bc} +\frac{a}{ab+ac} +\frac{b}{ab+bc} = \\ =\frac{ac*bc}{ac+bc}+\frac{ab*ac}{ab+ac}+\frac{ab*bc}{ab+bc} \leq \\ \leq \frac{ac+bc}{4}+ \frac{ab+ac}{4} + \frac{ab+bc}{4} = \frac{ab+bc+ac}{2} \leq \frac{a^2+b^2+c^2}{2}$     ultima relatie fiind binecunoscuta.