Lời giải:

\(A=4(a+b)^2+c^2-4c(a+b)+4(b+c)^2+a^2-4a(b+c)+4(c+a)^2+b^2-4b(a+c)\)

\(\Leftrightarrow A=4(a+b)^2+4(b+c)^2+4(c+a)^2-8(ab+bc+ac)\)

\(\Leftrightarrow A=4(a^2+b^2+2ab)+4(b^2+c^2+2bc)+4(c^2+a^2+2ac)-8(ab+bc+ac)\)

\(\Leftrightarrow A= 8(a^2+b^2+c^2)=8m\)

1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)

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

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)

có cả mấy bất đẳng thức đó hả

bn viết công thức tổng quát ra cho mk vs

mk thanks

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

\(A=\left(2a+2b+2c-3c\right)^2+\left(2b+2c+2a-3a\right)^2+\left(2c+2a+2b-3b\right)^2\)

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

Đặt \(a+b+c=n\)

\(\Rightarrow A=\left(2n-3c\right)^2+\left(2n-3a\right)^2+\left(2n-3b\right)\)

\(A=4n^2-12cn+9c^2+4n^2-12an+9a^2+4n^2-12bn+9b^2\)

\(A=12n.\left(n-a-b-c\right)+9.\left(a^2+b^2+c^2\right)\)

Ta có: \(a^2+b^2+c^2=m\)

\(\Rightarrow A=12.\left(a+b+c-a-b-c\right)+9m\)

\(A=9m\)

Vậy \(A=9m\)tại \(a^2+b^2+c^2=m\)

Tham khảo nhé~

Lời giải:

Đặt \(a+b+c=t\)

\(A=(2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2\)

\(=(2a+2b+2c-3c)^2+(2b+2c+2a-3a)^2+(2c+2a+2b-3b)^2\)

\(=(2t-3c)^2+(2t-3a)^2+(2t-3b)^2\)

\(=4t^2+9c^2-12tc+4t^2+9a^2-12ta+4t^2+9b^2-12tb\)

\(=12t^2+9(a^2+b^2+c^2)-12t(a+b+c)\)

\(=12t^2+9m-12t^2=9m\)

Áp dụng t/c dtsbn ta có:

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

\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)

\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)

\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)