Áp dụng bất đẳng thức Bunhiacopxki dạng phân thức ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)

Dấu = xảy ra khi a=b=c=1/3

Áp dụng hệ quả bất đẳng thức Cô - si , ta có :
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)
\(\Leftrightarrow\)\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\cdot1\ge9\)

\(\Leftrightarrow\)\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Áp dụng BĐT Cauchy Shwarz dạng Engel ta được:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \dfrac{(1+1+1)^2}{a+b+c}=\dfrac{9}{1}\)

\(\to \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge 9\)

\(\to\) Dấu "=" xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\)

\(\to a=b=c\)

Ta có:

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

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

Xét:

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

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

\(=1+\dfrac{c}{a-b}\left[\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab}\right]=1+\dfrac{c}{a-b}.\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}\)

\(=1-\dfrac{c\left(a+b-c\right)}{ab}=1-\dfrac{c.\left(-2c\right)}{ab}=1+\dfrac{2c^2}{ab}\) (do \(a+b+c=0\Rightarrow a+b=-c\))

Tương tự:

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

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

\(\Rightarrow P=3+2\left(\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\right)=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)

Mặt khác ta có đằng thức quen thuộc:

Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)

\(\Rightarrow P=3+\dfrac{2.3abc}{abc}=9\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}\geq \frac{(a+b+c)^2}{ab+bc+ac}\)

Ta sẽ cm \(\frac{(a+b+c)^2}{ab+bc+ac}\geq \frac{9}{a+b+c}\Leftrightarrow (a+b+c)^3\geq 9(ab+bc+ac)\)

\(\Leftrightarrow \sqrt{[3+2(ab+bc+ac)]^3}\geq 9(ab+bc+ac)\)

Đặt \(\sqrt{3+2(ab+bc+ac)}=t\) thì dễ thấy $0< t\leq 3$

Khi đó: 

\((a+b+c)^3\geq 9(ab+bc+ac)\Leftrightarrow t^3\geq 9.\frac{t^2-3}{2}\)

\(\Leftrightarrow 2t^3-9t^2+27\geq 0\)

$\Leftrightarrow (t-3)^2(2t+3)\geq 0$. Luôn đúng với mọi $t>0$

Vậy ta có đpcm

Dấu "=' xảy ra khi $a=b=c=1$

Áp dụng BĐT Cô -si cho 3 số dương:

\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\frac{\left(a+b\right)^3}{ab+9}+\frac{2}{3}\left(ab+9\right)+12\ge6a+6b\)

\(\Sigma\frac{a^3+b^3}{ab+9}\ge\frac{1}{4}\Sigma\frac{\left(a+b\right)^3}{ab+9}\ge\frac{1}{4}\left(12\left(a+b+c\right)-\frac{2}{3}\left(\frac{\left(a+b+c\right)^2}{3}+27\right)-36\right)=9\)