Đề sai. Bạn coi lại đề.

\(\dfrac{1}{a^2+a+1}\ge\dfrac{1}{a^2+\dfrac{a^2+1}{2}+1}=\dfrac{2}{3}.\dfrac{1}{a^2+1}=\dfrac{2}{3}\left(1-\dfrac{a^2}{a^2+1}\right)\ge\dfrac{2}{3}\left(1-\dfrac{a}{2}\right)\)

Tương tự và cộng lại: \(VT\ge\dfrac{2}{3}\left(3-\dfrac{a+b+c}{2}\right)=\dfrac{2}{3}.\dfrac{3}{2}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{bc+ac+ab}{abc}=0\)

\(\Leftrightarrow ab+bc+ca=0\) (*)

Lại có: \(a+b+c=1\Leftrightarrow\left(a+b+c\right)^2=1^2\)

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

Kết hợp với (*) \(\Leftrightarrow a^2+b^2+c^2=1\)(đpcm)

Ta có: \(a+b+c=1\)

\(\Rightarrow\left(a+b+c\right)^2=1\)

\(\Rightarrow a^2+b^2+c^2+2.\left(ab+bc+ca\right)=1\left(1\right)\)

Lại có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Rightarrow2.\left(ab+bc+ca\right)=0\left(2\right)\) ( Nhân 2 vế cho 2abc khác 0 )

Lấy \(\left(1\right)\) trừ \(\left(2\right)\) vế theo vế ta được \(a^2+b^2+c^2=1\)

\(\Rightarrow\) Đpcm.

co ai ko giup

co ai   ko

* Áp dụng BĐT \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ vào bài toán có :

\(\dfrac{1}{4}\cdot\left(\dfrac{4}{a+b}\right)\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế với vế các BĐT có :

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

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

Giả sử \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\Rightarrow ac-ab=ab-bc\Rightarrow ac+bc=2ab\left(\text{luôn đúng}\right)\)

Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)