ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Rightarrow\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)

\(\Rightarrow\frac{2}{c}=\frac{b+a}{ab}\)

\(\Rightarrow2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\)

\(\Rightarrow ac-ab=ab-bc\)

\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)

tíc mình nha

a) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\) 

  \(a^2+b^2+c^2+2ab+2ac+2bc-3ab-3ac-3bc=0\) 

 \(a^2+b^2+c^2-ab-ac-bc=0\) 

\(2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\) 

 \(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\) 

\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\) 

\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\) 

\(\Rightarrow a=b=c\left(đpcm\right)\)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}=\dfrac{1}{c\left(a^2+b^2\right)}+\dfrac{1}{a\left(b^2+c^2\right)}+\dfrac{1}{b\left(c^2+a^2\right)}\)

\(\ge\dfrac{9}{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}\ge\dfrac{9}{2\left(a^3+b^3+c^3\right)}\)

\(\Rightarrow P\ge a^3+b^3+c^3+\dfrac{9}{2\left(a^3+b^3+c^3\right)}\ge3\sqrt[3]{\left(\dfrac{a^3+b^3+c^3}{2}\right)^2.\dfrac{9}{2\left(a^3+b^3+c^3\right)}}\)

\(=3\sqrt[3]{\dfrac{9\left(a^3+b^3+c^3\right)}{8}}\ge3\sqrt[3]{\dfrac{27abc}{8}}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đáp án:

Cho a,b,c thỏa mãn:

2ab(2b-a)-2ac(c-2a)-2bc(b-2c)= 7abc

CMR:Tồn tại 1số bằng 2 số kia.

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