\(\Leftrightarrow a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\)

\(\Leftrightarrow ab=2c^2+ca\Leftrightarrow ab+ac=2c^2+2ac\)

\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\rightarrowđpcm\)

\(\hept{\begin{cases}a^2+ab+\frac{b^2}{3}=25\\c^2+\frac{b^2}{3}=9\end{cases}}\Rightarrow a^2+ac-c^2=16\)

\(\Rightarrow a^2+ab-c^2=a^2+ac+c^2\left(=16\right)\)

\(\Rightarrow ab-c^2=ac+c^2\)

\(\Rightarrow ab=ac+2c^2\)

\(\Rightarrow ab+ac=2ac+2c^2\)

\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\)

\(\Leftrightarrow\frac{2c}{a}=\frac{b+c}{a+c}\left(đpcm\right)\)

\(A=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)

\(A< \frac{1}{100\cdot101}+\frac{1}{101\cdot102}+\frac{1}{102\cdot103}+\frac{1}{103\cdot104}+\frac{1}{104\cdot105}\)

\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)

\(=\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{2^2\cdot3\cdot5^2\cdot7}=B\)

Vậy \(A< B\)

Có \(a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\left(=25\right)\)

\(\Rightarrow a^2+ab+\frac{b^2}{3}=2c^2+\frac{b^2}{3}+a^2+ac\\ \Rightarrow ab=2c^2+ac\\ \Rightarrow ab+ac=2c^2+2ac\\ \Rightarrow a\left(b+c\right)=2c\left(a+c\right)\\ \Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\)

A = \(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)< \(\frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\) =\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\) 

= \(\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}\)= \(\frac{1}{2^2.3.5^2.7}\)= B

Vậy A < B

\(A