Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)
Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)
\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)
Áp dụng t/c bắc cầu ta dc : \(a=b=c\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)
=>a2+b2=2ab
=>a2-2ab+b2=0
=>(a-b)2=0=>a=b
tương tự=>b=c
=>a=b=c
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)
Ta có:
\(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2+b^2-2ab=0\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a-b=0\)
\(\Rightarrow a=b\left(1\right)\)
\(\frac{b^2+c^2}{2}=bc\Rightarrow b^2+c^2=2bc\)
\(\Rightarrow b^2+c^2-2bc=0\)
\(\Rightarrow\left(b-c\right)^2=0\)
\(\Rightarrow b-c=0\)
\(\Rightarrow b=c\left(2\right)\)
\(\frac{a^2+c^2}{2}=ac\Rightarrow a^2+c^2=2ac\)
\(\Rightarrow a^2+c^2-2ac=0\)
\(\Rightarrow\left(a-c\right)^2=0\)
\(\Rightarrow a-c=0\)
\(\Rightarrow a=c\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow a=b=c\)
\(\Rightarrow a+b+c=3a\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3.\frac{1}{a}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3.\frac{1}{a}=9\)
Vậy \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=9\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)\(=1+\frac{a}{b}+\frac{b}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
\(=\left(1+1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
\(=3+\frac{a^2+b^2}{ab}+\frac{a^2+c^2}{ac}+\frac{b^2+c^2}{bc}\)= \(3+\frac{a^2+b^2}{\frac{a^2+b^2}{2}}+\frac{a^2+c^2}{\frac{a^2+c^2}{2}}+\frac{b^2+c^2}{\frac{b^2+c^2}{2}}\)= 3 + 2 + 2 + 2 = 9