\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)

\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)

\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

Lời giải:

Xét tử :

\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}=\frac{a^2}{a^2+bc+(-ab-ac)}+\frac{b^2}{b^2+ac+(-ab-bc)}+\frac{c^2}{c^2+ab+(-bc-ac)}\)

\(=\frac{a^2}{a(a-b)-c(a-b)}+\frac{b^2}{b(b-c)-a(b-c)}+\frac{c^2}{c(c-a)-b(c-a)}\)

\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\)

\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Xét mẫu (tương tự bên tử)

\(\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}=\frac{bc}{(a-c)(a-b)}+\frac{ac}{(b-a)(b-c)}+\frac{ab}{(c-a)(c-b)}\)

\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Do đó:

\(A=\frac{1}{1}=1\)

\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)

\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)

Áp dụng BĐT Cauchy ta có

\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)

\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)

\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh

\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)

\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)

Áp dụng BĐT Cauchy:

\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)

\(a+b+c\ge3\sqrt[3]{abc}=3\)

\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm

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

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

\(C=\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)

\(=\dfrac{a^2}{a^2+bc-ac-ab}+\dfrac{b^2}{b^2+ac-ba-bc}+\dfrac{c^2}{c^2+ab-ca-cb}\)

\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=-\left(\dfrac{a^2}{\left(a-b\right)\left(c-a\right)}+\dfrac{b^2}{\left(a-b\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)

\(=-\left(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)

\(=-\left(\dfrac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)=1\)

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

Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(a-c\right)\left(b-c\right)\)

\(\Rightarrow A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}\)

\(\Rightarrow A=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

Biến đổi tử số:

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

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

Vậy \(A=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Ta có : 1/M=a²+2bc+b²+2ac+c²+2ab

=(a+b+c)² ➝ M=1/(a+b+c)²

mik nghĩ là thế

Có:

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

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

\(\Leftrightarrow ab+bc+ac=0\)

\(1\Leftrightarrow a^2+2bc=a^2+bc-ab-ac\)

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

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

\(2\Leftrightarrow b^2+2ac=b^2+ac-ab-bc\)

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

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

\(3.c^2+2ab=c^2+ab-bc-ac\)

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

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

\(\Rightarrow M=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)

\(\Rightarrow M=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)

\(\Rightarrow M=\dfrac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\Rightarrow M=0\)

Chắc chắn là đề bài thiếu rồi

1. Thiếu điều kiện liên quan a;b;c (là số dương hay số gì)

2. Thiếu mối liên hệ giữa a;b;c (a;b;c bất kì thì BĐT này hiển nhiên sai)