Inegalitatea mediilor:
[tex]m_h \leq m_g \leq m_a\leq m_p[/tex]
Unde [tex]\displaystyle m_h(x,y) = \frac{2xy}{x+y}[/tex] și [tex]\displaystyle m_a(x,y) = \frac{x+y}{2}[/tex]
Mă voi folosi doar de [tex]m_h \leq m_a[/tex] :
[tex]\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[/tex]
[tex]\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}[/tex]
[tex]\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)[/tex]
[tex]\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)[/tex]
[tex]\Rightarrow \displaystyle \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{a+c}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/tex]
[tex]\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)}[/tex]
