\(a^2+b^2+c^2=ab+bc+ca\\ \Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\ \Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ Vì\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\in R\\\left(b-c\right)^2\ge0\forall b,c\in R\\\left(c-a\right)^2\ge0\forall c,a\in R\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\\ \Rightarrow a=b=c\\ Khiđó:A=0\)

(a+b+c/3)²= a²+b²+(c/3)²+2ab+2/3ac+2/3bc

* a²+b²+(c/3)² \(\ge\)0

=> a²+b²+(c/3)²+2ab+2/3ac+2/3bc\(\ge\)2ab+2/3ac+2/3bc

mà 2ab+2/3ac+2/3bc\(\ge\)ab+bc+ca

=> a²+b²+(c/3)²+2ab+2/3ac+2/3bc\(\ge\)ab+bc+ca

=> (a+b+c/3)²\(\ge\)ab+bc+ca

trả lời:

(a+b+c/3)2= a2+b2+(c/3)2+2ab+2/3ac+2/3bc

* a2+b2+(c/3)2 \ge≥0

=> a2+b2+(c/3)2+2ab+2/3ac+2/3bc\ge≥2ab+2/3ac+2/3bc

mà 2ab+2/3ac+2/3bc\ge≥ab+bc+ca

=> a2+b2+(c/3)2+2ab+2/3ac+2/3bc\ge≥ab+bc+ca

=> (a+b+c/3)2\ge≥ab+bc+ca

\(A=a+b+c+\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\dfrac{1}{2}\left(a+b+c\right)^2+\left(a+b+c\right)-\dfrac{3}{2}\)

\(A=\dfrac{1}{2}\left(a+b+c+1\right)^2-2\ge-2\)

\(A_{min}=-2\) khi \(a+b+c=-1\) (có vô số bộ a;b;c thỏa mãn điều này)

Với mọi a;b;c ta luôn có:

\(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+3\ge2\left(a+b+c+ab+bc+ca\right)\)

\(\Leftrightarrow12\ge2A\)

\(\Rightarrow A\le6\)

\(A_{max}=6\) khi \(a=b=c=1\)

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)

\(=12\left(x-y\right)^2=12.12^2=1728\)

Đề bài sai

Phản ví dụ: với \(a=b=c=-2\) thì \(a^2+b^2+c^2+abc+4< 2\left(ab+bc+ca\right)\)

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

\(=\dfrac{a^3+a^2b+a^2c+ab^2+b^2+b^2c+ac^2+bc^2+c^3+a^2b^2+b^2c^2+a^2c^2+2ab^2c+2a^2bc+2abc^2}{a^2+b^2+c^2+2ab+2bc+2ac-\left(ab+bc+ac\right)}\)

\(=\dfrac{a^3+a^2b+a^2c+ab^2+b^3+b^2c+ac^2+bc^2+c^3+a^2b^2+b^2c^2+a^2c^2-2ab^2c+2a^2bc+2abc^2}{a^2+b^2+c^2+ab+ac+bc}\)