a) \(4\left(a+b\right)ab=3\left(a-b\right)^2+\left(a+b\right)^2\Leftrightarrow4\left(a+b\right)ab=4a^2+4b^2-4ab\Leftrightarrow\left(a+b\right)ab=a^2+b^2-ab\) (đúng)

=> đẳng thức được cm

b) nếu nghĩ ra thì tớ giải cho

Trước hết theo BĐT Schur bậc 3 ta có:

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)

Đặt vế trái BĐT cần chứng minh là P, ta có:

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

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

Áp dụng (1):

\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(Ta\) \(có:\) \(1+a^2=ab+bc+ca+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(c+a\right)\)

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

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

\(Khi\) \(đó:\) \(A=\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(c+b\right)}\)

\(\Rightarrow A=1\)

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

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

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

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

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

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

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

Dấu "=" xảy ra khi \(a=b=c=1\)

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow x+y+z=3\)

\(K=\dfrac{z^3}{x^2+z^2}+\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}\)

Ta chứng minh BĐT phụ sau: \(\dfrac{x^3}{x^2+y^2}\ge\dfrac{2x-y}{2}\)

Thật vậy, BĐT tương đương:

\(2x^3\ge2x^3-x^2y+2xy^2-y^3\)

\(\Leftrightarrow y\left(x-y\right)^2\ge0\) (đúng)

Tương tự: \(\dfrac{y^3}{y^2+z^2}\ge\dfrac{2y-z}{2}\) ; \(\dfrac{z^3}{z^2+x^2}\ge\dfrac{2z-x}{2}\)

Cộng vế với vế:

\(K\ge\dfrac{x+y+z}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=\dfrac{1}{3}\)

Ta có: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)

Tương tự: \(\left\{{}\begin{matrix}b^2+1=\left(a+b\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)

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

Mặt khác: \(a+b+c-abc=a\left(1-bc\right)+b+c\)

                \(=a\left(ab+ca\right)+b+c\)     (Vì ab+bc+ca=1)

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

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

\(T=1\)

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:

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

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

\(a^2+b^2+c^2\ge ab+bc+ca=2\)

Áp dụng BĐT C-S:

\(P\ge\dfrac{\left(a+b+c\right)^2}{3-\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2+4}{3-\left(a^2+b^2+c^2\right)}\)

Đặt \(a^2+b^2+c^2=x\)

Ta cần c/m: \(\dfrac{x+4}{3-x}\ge6\Leftrightarrow x+4\ge18-6x\)

\(\Leftrightarrow x\ge2\) (đúng)

Dấu = xảy ra khi \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)

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