Ta có:\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a}{b}\cdot\frac{b}{c}=\frac{a}{c}\)

Mà\(\frac{a}{b}=\frac{b}{c}=\frac{2015b}{2015c}=\frac{a+2015b}{b+2015c}\)

Nên suy ra\(\frac{a}{c}=\frac{a^2}{b^2}=\left(\frac{a+2015b}{b+2015c}\right)^2=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\)

           Vậy\(\frac{a}{c}=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\left(đpcm\right)\)

B1:

Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)

Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:

\(2b=a+\frac{2bd}{b+a}\)

\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)

\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)

\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)

\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)

\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)

B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z