Lời giải:

Đặt \(a+b+c=t\)

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

Đặt a + b + c = x thì:

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

\(=4x^2-12cx+9c^2+4x^2-12ax+9a^2+4x^2-12bx+9b^2\)

\(=12x^2-12x\left(a+b+c\right)+9\left(a^2+b^2+c^2\right)\)

\(12x^2-12x^2+9\left(a^2+b^2+c^2\right)=9\left(a^2+b^2+c^2\right)=9m\)

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

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

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

\(a+b=1-c>\frac{1}{2}>c\)

Tương tự \(b+c>a;a+c>b\)

\(VT=\frac{1}{a\left(b+c-a\right)}+\frac{1}{b\left(a+c-b\right)}+\frac{1}{c\left(a+b-c\right)}\)

\(VT\ge\frac{4}{\left(a+b+c-a\right)^2}+\frac{4}{\left(b+a+c-b\right)^2}+\frac{4}{\left(c+a+b-c\right)^2}\)

\(VT\ge\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\ge\frac{4}{3}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)^2\)

\(VT\ge\frac{4}{3}\left(\frac{9}{2\left(a+b+c\right)}\right)^2=\frac{4.81}{3.4}=27\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

1.

\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)

Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)

Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)

Từ đó ta được đpcm

2.

\(a,Sửa:a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^6-b^6\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^2-b^2+1\right)\\ =\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^2-b^2+1\right)\\ b,=\left(a^3+b^3\right)-1+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\\ =\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\\ =\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\)

\(c,=a^2b^2\left(b-a\right)+b^2c^2\left(c-a+a-b\right)-c^2a^2\left(c-a\right)\\ =-a^2b^2\left(a-b\right)+b^2c^2\left(a-b\right)+b^2c^2\left(c-a\right)-c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(b^2c^2-a^2b^2\right)+\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(c-a\right)\left(c+a\right)+c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(c-a\right)\left[b^2\left(c+a\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b^2c+ab^2-bc^2-ac^2\right)\\ =\left(a-b\right)\left(c-a\right)\left[bc\left(b-c\right)+a\left(b-c\right)\left(b+c\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(bc+ab+ac\right)\)

Từ \(a^2b^2+b^2c^2+c^2a^2\ge a^2b^2c^2\)\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\)

bài này tui làm rồi ở đây

\(8VT=4\left(a^2b+b^2c+c^2a+abc\right)\left(2ab^2+2bc^2+2ca^2+2abc\right)\le\left(a^2b+b^2c+c^2a+2ab^2+2bc^2+2ca^2+3abc\right)^2\)

\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+6abc\right)^2\)

\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+9abc\right)^2\)

\(\Rightarrow VT\le\frac{1}{32}\left[\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\right]^2\)

\(\Rightarrow VT\le\frac{1}{512}\left[\left(a+2b\right)\left(4b+8c\right)\left(c+2a\right)\right]^2\)

\(\Rightarrow VT\le\frac{1}{512}\left(\frac{a+2b+4b+8c+c+2a}{3}\right)^6=\frac{1}{512}\left(a+2b+3c\right)^6=\frac{4^6}{512}=8\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;1;0\right)\)