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

\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

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

Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)

Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)

\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)

\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)

\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\) 

chúc bạn học tốt ^ ^

Ta có:

\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

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

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

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

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

\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

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

Đồng thời:

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

Tương tự:

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

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

Từ đó:

\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)

Từ giả thiết ta có:

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

\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)

Tương tự:

\(b^2-c^2-a^2=2ca,c^2-a^2-b^2=2ab\)

Từ đây suy ra:

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

Mặt khác lại có:

\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^3=-a^3\)

\(\Rightarrow b^3+c^3+3bc\left(b+c\right)=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow A=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3}{2}\)

Ngô Tấn Đạt

Ngô Thanh Sang

Ta có: a+b+c=0

<=>a=-b-c

<=>\(a^2=(-b-c)^2\)

<=>\(a^2=b^2+c^2+2bc\)

cmtt:\(b^2=a^2+c^2+2ac\)

\(c^2=a^2+b^2+2ab\)

=> A=\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3}{2abc}+\frac{b^3}{ 2abc}+\frac{c^3}{2abc} \)

=\(\frac{1}{2abc}(a^3+b^3+c^ 3)\)

Cm đẳng thức phụ

Với a+b+c=0=> \(a^3+b^3+c^3=3abc\)

=>A=\(\frac{3}{2} \)

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

\(M\ge\dfrac{\left(ab+bc+ca\right)^2}{2abc\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2abc}=\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{6abc}\ge\dfrac{9abc}{6abc}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=2(\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2})+\frac{1}{2(ab+bc+ac)}\\ \geq 2.\frac{9}{2(ab+bc+ac)+a^2+b^2+c^2}+\frac{1}{2(ab+bc+ac)}\\ =\frac{18}{(a+b+c)^2}+\frac{1}{2(ab+bc+ac)}\\ =18+\frac{1}{2(ab+bc+ac)}\)

Áp dụng BĐT AM-GM:

$2(ab+bc+ac)\leq 2.\frac{(a+b+c)^2}{3}=\frac{2}{3}$

$\Rightarrow \frac{1}{2(ab+bc+ac)}\geq \frac{3}{2}$

$\Rightarrow P\geq 18+\frac{3}{2}=\frac{39}{2}$
Vậậy $P_{\min}=\frac{39}{2}$ khi $a=b=c=\frac{1}{3}$

\(\sqrt{ab}+\sqrt{4b.c}+2\left(a+c\right)\le\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(4b+c\right)+2\left(a+c\right)=\dfrac{5}{2}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{2}{5}\left(\dfrac{1}{a+b+c}-\dfrac{1}{\sqrt{a+b+c}}\right)=\dfrac{2}{5}\left(\dfrac{1}{\sqrt{a+b+c}}-\dfrac{1}{2}\right)^2-\dfrac{1}{10}\ge-\dfrac{1}{10}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a+b+c=4\\a=b=\dfrac{c}{4}\end{matrix}\right.\) em tự giải ra a;b;c