Question

Fie a,b numere reale, pozitive. Sa se arate ca $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}$.

Answer 1

$\frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{a+b} \geq \frac{3}{2} \Leftrightarrow \\ \\ \Leftrightarrow \frac{a^2}{a(b+c)}+ \frac{b^2}{b(a+c)}+ \frac{c^2}{c(a+b)} \geq \frac{3}{2} . \\ \\ Folosind~indegalitatea~Cauchy-Buniakovski-Schwartz~obtinem: \\ \\ \frac{a^2}{a(b+c)}+ \frac{b^2}{b(a+c)}+ \frac{c^2}{c(a+b)} \geq \frac{(a+b+c)^2}{a(b+c)+b(a+c)+c(a+b)}= \frac{(a+b+c)^2}{2ab+2bc+2ac}. \\ \\ Ramane~sa~verificam~daca~ \frac{(a+b+c)^2}{2ab+2bc+2ac} \geq \frac{3}{2}.$

$\frac{(a+b+c)^2}{2ab+2bc+2ac} \geq \frac{3}{2} \Leftrightarrow \\ \\ \Leftrightarrow 2(a+b+c)^2 \geq 3(2ab+2bc+2ac) \Leftrightarrow \\ \\ \Leftrightarrow 2(a^2+b^2+c^2+2ab+2bc+2ac) \geq 6ab+6bc+6ac \Leftrightarrow \\ \\ \Leftrightarrow 2a^2+2b^2+2c^2+4ab+4bc+4ac \geq 6ab+6bc+6ac \Leftrightarrow \\ \\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac \geq 0 \Leftrightarrow \\ \\ \Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2) \geq 0 \Leftrightarrow$

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2 \geq 0,~evident! \\ \\ Astfel~am~incheiat~demonstratia.~Egalitatea~are~loc~pnetru \\ \\ a=b=c.$

Answer 2

 Notam b+c=x, a+c=y si a+b=z =>
a=(y+z-x)/2
b=(x+z-y)/2
c=(x+y-z)/2
Inegalitatea se transforma in:
[tex] \frac{y+z-x}{2x} + \frac{x+z-y}{2y}+ \frac{x+y-z}{2z}\geq \frac{3}{2} |\cdot 2\\ \frac{y+z-x}{x} + \frac{x+z-y}{y}+ \frac{x+y-z}{z}\geq 3\\ \frac{y}{x} +\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1+\frac{x}{z}+\frac{y}{z}-1\geq3\\ ( \frac{x}{y} + \frac{y}{x} )+( \frac{x}{z}+ \frac{z}{x} } )+( \frac{y}{z} + \frac{z}{y})\geq6(Adevarata) \ deoarece :\\ \frac{x}{y} + \frac{y}{x}\geq2\\ \frac{x}{z}+ \frac{z}{x} \geq2\\ \frac{y}{z} + \frac{z}{y}\geq2[/tex]