Ta có: \(\frac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\)\(=\frac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)

Ta luôn có: \(\left(\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right)^2\ge0\)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2+2.\)\(\left(\frac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\right)\ge0\)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2\ge2\)(*)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+1+\left(\frac{b+c}{b-c}\right)^2+1+\left(\frac{c+a}{c-a}\right)^2+1\ge5\)

\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{\left(a-b\right)^2}+\frac{2\left(b^2+c^2\right)}{\left(b-c\right)^2}+\frac{2\left(c^2+a^2\right)}{\left(c-a\right)^2}\ge5\)\(\Leftrightarrow\frac{a^2+b^2}{\left(a-b\right)^2}+\frac{b^2+c^2}{\left(b-c\right)^2}+\frac{c^2+a^2}{\left(c-a\right)^2}\ge\frac{5}{2}\)(1)

(*)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2-1+\left(\frac{b+c}{b-c}\right)^2-1+\left(\frac{c+a}{c-a}\right)^2-1\ge-1\)\(\Leftrightarrow\frac{4ab}{\left(a-b\right)^2}+\frac{4bc}{\left(b-c\right)^2}+\frac{4ca}{\left(c-a\right)^2}\ge-1\)\(\Leftrightarrow\frac{ab}{\left(a-b\right)^2}+\frac{bc}{\left(b-c\right)^2}+\frac{ca}{\left(c-a\right)^2}\ge-\frac{1}{4}\)(2)

Lấy (1) + (2), ta được: \(\frac{a^2+ab+b^2}{\left(a-b\right)^2}+\frac{b^2+bc+c^2}{\left(b-c\right)^2}+\frac{c^2+ca+a^2}{\left(c-a\right)^2}\ge\frac{9}{4}\)

\(\Leftrightarrow\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)(đpcm)

Chú ý: Từ đây ta có thể biến thành một BĐT khác khó hơn: \(\frac{a^3+b^3}{\left(a-b\right)^3}+\frac{b^3+c^3}{\left(b-c\right)^3}+\frac{c^3+a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)

Ta có: \(\sqrt{2x^2-4x+5}=\sqrt{2x^2-4x+2+3}=\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\)

Lại có: \(\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(\sqrt{2}x-\sqrt{2}\right)^2+3\ge3\)

\(\Rightarrow\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\ge\sqrt{3}\)

Vậy Min y là \(2+\sqrt{3}\)

\(y=2+\sqrt{2x^2-4x+5}=2+\sqrt{2x^2-4x+2+3}\)

\(=2+\sqrt{2\left(x^2-2x+1\right)+3}=2+\sqrt{2\left(x-1\right)^2+3}\)

Vì \(\left(x-1\right)^2\ge0\)\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2\ge0\)\(\forall x\)\(\Rightarrow2\left(x-1\right)^2+3\ge3\)\(\forall x\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\)\(\forall x\)

\(\Rightarrow y=2+\sqrt{2\left(x-1\right)^2+3}\ge2+\sqrt{3}\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(miny=2+\sqrt{3}\)\(\Leftrightarrow x=1\)

- Vì vai trò của a , b ,c trong bài này là như nhau nên có thể giả sử \(a\le b\le c\)mà không làm giảm đi tính tổng quát của bài toán . Khi đó ta có :

\(3=a+b+c\le3c\Rightarrow c\ge1\Rightarrow1\le x\le2\)

Ta có : \(a^2+b^2\le\left(a+b\right)^2\)(vì \(a,b\ge0\))

\(\Rightarrow A\le\left(a+b\right)^2+c^2=\left(3-c\right)^2+c^2=2c^2-6c+9\)

          \(\le2.\left(c^2-3c+\frac{9}{4}\right)+\frac{9}{2}=2\left(c-\frac{3}{2}\right)^2+\frac{9}{2}\)

Do \(1\le c\le2\)nên \(-\frac{1}{2}\le x-\frac{3}{2}\le\frac{1}{2}\Rightarrow|c-\frac{3}{2}|\le\frac{1}{2}\)

\(\Rightarrow2|x-\frac{3}{2}|^2+\frac{9}{2}\le2.\frac{1}{4}+\frac{9}{2}=5\Rightarrow A\le5\)

Dễ thấy khi a = 0 ; b = 1 ; c = 2 thỏa mãn \(a,b,c\in\left[0;2\right];a+b+c=3\)và \(a\le b\le c\)thì A = 5

Vậy : \(A_{max}=5\)

Do \(a,b,c\in\left[0;2\right]\)nên \(\left(a-2\right)\left(b-2\right)\left(c-2\right)\le0\)\(\Leftrightarrow abc-2\left(ab+bc+ca\right)+4\left(a+b+c\right)-8\le0\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4+abc\Leftrightarrow\left(a+b+c\right)^2\ge a^2+b^2+c^2+abc+4\)\(\Leftrightarrow a^2+b^2+c^2\le5-abc\le5\)(Do \(a,b,c\ge0\))

Đẳng thức xảy ra khi trong 3 số a, b, c có một số bằng 0, một số bằng 1 và một số bằng 2