\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ac}{a+c}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{a+c}{ac}=\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\left(vì\text{ a;b;c dương}\right)\)

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

a² + b² + c² = ( a - b )² + ( b - c )² + ( c - a )²

<=> a² + b² + c² = a² - 2ab + b² + b² - 2bc + c² + c² - 2ca + a²

<=> a² + b² + c² - 2ab - 2bc - 2ca = 0 ( bớt a² + b² + c² ở cả hai vế )

<=> a² + b² + c² - 2( ab + bc + ca ) = 0

<=> a² + b² + c² - 2.9 = 0

<=> a² + b² + c² - 18 = 0

<=> a² + b² + c² = 18

Xét ( a + b + c )² ta có :

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca 

                     = ( a² + b² + c² ) + 2( ab + bc + ca )

                     = 18 + 2.9

                     = 18 + 18 = 36

=> ( a + b + c )² = 36

=> a + b + c = 6 ( do a, b, c là các số dương )

\(a^2+ab+b^2=\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{1}{2}\left(a+b\right)^2\ge\dfrac{1}{4}\left(a+b\right)^2+\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{4}\left(a+b\right)^2\)

\(\Rightarrow\sqrt{a^2+ab+b^2}\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2}=\dfrac{\sqrt{3}}{2}\left(a+b\right)\)

Tương tự và cộng lại:

\(P\ge\sqrt{3}\left(a+b+c\right)=\sqrt{3}\)

\(P_{min}=\sqrt{3}\) khi \(a=b=c=\dfrac{1}{3}\)

\(\sqrt{a^2-ab+b^2}=\sqrt{\left(a+b\right)^2-3ab}\ge\sqrt{\left(a+b\right)^2-\frac{3\left(a+b\right)^2}{4}}=\sqrt{\frac{\left(a+b\right)^2}{4}}=\frac{\left(a+b\right)}{2}\)

tượng tự...
A>=3

Gọi \(S=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+ab+c^2}+\frac{a^3}{c^2+ab+a^2}\)

Dễ thấy \(P-S=0\)

\(\Rightarrow2P=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+ab+c^2}+\frac{c^3+a^3}{c^2+ab+a^2}\)

Ta chứng minh: 

\(\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{a+b}{3}\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)(đúng)

\(\Rightarrow2P\ge\frac{a+b}{3}+\frac{b+c}{3}+\frac{c+a}{3}=\frac{2\left(a+b+c\right)}{3}=2\)

\(\Rightarrow P\ge1\)

P-S=0 ?? =))