Question

Demonstrati ca $\frac{1}{a+b}$ + $\frac{1}{b+c}$ + $\frac{1}{a+c}$ $\leq$ $\frac{1}{2}$ ( $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ ), oricare ar fi a, b, c numere reale pozitive.

Answer 1

Inegalitatea mediilor:

$m_h \leq m_g \leq m_a\leq m_p$

Unde  $\displaystyle m_h(x,y) = \frac{2xy}{x+y}$  și  $\displaystyle m_a(x,y) = \frac{x+y}{2}$

Mă voi folosi doar de  $m_h \leq m_a$ :

$\left.\begin{cases}\displaystyle \frac{2\cdot \frac{1}{a}\cdot \frac{1}{b}}{\frac{1}{a}+\frac{1}{b}} \leq \frac{\frac{1}{a}+\frac{1}{b}}{2} \\ \\\displaystyle\frac{2\cdot \frac{1}{b}\cdot \frac{1}{c}}{\frac{1}{b}+\frac{1}{c}} \leq \frac{\frac{1}{b}+\frac{1}{c}}{2}\\ \\\displaystyle \frac{2\cdot \frac{1}{a}\cdot \frac{1}{c}}{\frac{1}{a}+\frac{1}{c}} \leq \frac{\frac{1}{a}+\frac{1}{c}}{2}\end{cases}\right)(+) \,\,\,\Rightarrow$

$\Rightarrow \displaystyle \!{^{^{^{^{^{^{\displaystyle ab)}}}}}}}\!\!\frac{2\cdot \frac{1}{a}\cdot \frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\,+ \!\!{^{^{^{^{^{^{\displaystyle bc)}}}}}}}\!\!\frac{2\cdot \frac{1}{b}\cdot \frac{1}{c}}{\frac{1}{b}+\frac{1}{c}}\,+ \!{^{^{^{^{^{^{\displaystyle ac)}}}}}}}\!\!\frac{2\cdot \frac{1}{a}\cdot \frac{1}{c}}{\frac{1}{a}+\frac{1}{c}}\leq \frac{\frac{1}{a}+\frac{1}{b}}{2}+\frac{\frac{1}{b}+\frac{1}{c}}{2}+\frac{\frac{1}{a}+\frac{1}{c}}{2}$

$\Rightarrow \displaystyle \frac{2}{b+a}+\frac{2}{c+b}+\frac{2}{c+a}\leq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\dfrac{1}{c}\right)$

$\Rightarrow \displaystyle \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{a+c} \leq \frac{1}{2}\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\right)$

$\Rightarrow \displaystyle \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{a+c}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

$\Rightarrow \boxed{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\leq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$