[tex]\displaystyle\it\\a^2,b^2,c^2~sunt~in~progresie~aritmetica~\implies b^2=\frac{a^2+c^2}{2}.\\\frac{1}{a+b},\frac{1}{a+c},\frac{1}{b+c}~sunt~in~progresie~aritmetica \implies\\\frac{1}{a+c}=\frac{\frac{1}{a+b}+\frac{1}{b+c}}{2} =\frac{\frac{b+c+a+b}{(a+b)(b+c)}}{2}=\frac{\frac{2b+a+c}{(a+b)(b+c)}}{2}=\frac{2b+a+c}{2(a+b)(b+c)}=\\\frac{2b+a+c}{2ab+2ac+2b^2+2bc} \Leftrightarrow 2ab+2ac+2bc+2b^2=(a+c)(2b+a+c)\Leftrightarrow\\\\2ab+2ac+2bc+2b^2=2ab+a^2+2ac+2bc +c^2 \Leftrightarrow 2b^2=a^2+c^2 \Leftrightarrow[/tex]
[tex]\displaystyle\it\\b^2=\frac{a^2+c^2}{2},~dar~deja~a^2,b^2~si~c^2~sunt~in~progresie~aritmetica\\\implies \frac{1}{a+b},~\frac{1}{a+c},~\frac{1}{b+c}~sunt~in~progresie~aritmetica[/tex]
