Ta chứng minh bổ đề:

Với x,y,z dương thì:

\(8\left(x+y+z\right)\left(xy+yz+zx\right)\le9\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

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

Quay lại bài toán ta có:

\(A^{2020}=\left(\sqrt[2020]{\frac{a}{a+b}}+\sqrt[2020]{\frac{b}{b+c}}+\sqrt[2020]{\frac{c}{c+a}}\right)^{2020}\)

\(=\left(\sqrt[2020]{\frac{a\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}}+\sqrt[2020]{\frac{b\left(b+a\right)}{\left(b+c\right)\left(b+a\right)}}+\sqrt[2020]{\frac{c\left(c+b\right)}{\left(c+a\right)\left(c+b\right)}}\right)^{2020}\)

\(\le\left(1+1+1\right)^{2018}.2.\left(a+b+c\right).\left(\frac{a}{\left(a+b\right)\left(a+c\right)}+\frac{b}{\left(b+c\right)\left(b+a\right)}+\frac{c}{\left(c+a\right)\left(c+b\right)}\right)\)

\(=3^{2018}.\frac{4\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\le3^{2018}.\frac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{3^{2020}}{2}\)

\(\Rightarrow A\le\frac{3}{\sqrt[2020]{2}}\)

Đặt: f(a;b;c) =\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Vai trò của a, b, c là như nhau có thể giả sử: \(a=max\left\{a,b,c\right\}\)

Ta có: \(f\left(a;b;\sqrt{ab}\right)=\frac{a}{a+b}+\frac{b}{b+\sqrt{ab}}+\frac{\sqrt{ab}}{\sqrt{ab}+a}\)

\(=\frac{a}{a+b}+\frac{\sqrt{b}}{\sqrt{b}+\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{b}+\sqrt{a}}=\frac{a}{a+b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

Ta chứng minh:

\(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

+) Chứng minh: \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\)

Xét : \(f\left(a;b;c\right)-f\left(a;b;\sqrt{ab}\right)=\frac{b}{b+c}+\frac{c}{a+c}-\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

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

\(=\frac{ab\sqrt{a}-ab\sqrt{b}+2bc\sqrt{a}-2ac\sqrt{b}+c^2\sqrt{a}-c^2\sqrt{b}}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\ge0\)vì a=max{a,b,c} => \(a\ge b\)

=> \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\)(1)

+) Chứng minh:\(f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

Xét: \(f\left(a;b;\sqrt{ab}\right)-\frac{7}{5}=\frac{a}{a+b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{7}{5}\)\(=\frac{\frac{a}{b}}{\frac{a}{b}+1}+\frac{2}{\sqrt{\frac{a}{b}}+1}-\frac{7}{5}\)(2)

Đặt \(\sqrt{\frac{a}{b}}=x\left(đk:x\le3\right)\)Ta có: 

(2)=\(\frac{x^2}{x^2+1}+\frac{2}{x+1}-\frac{7}{5}\)\(=\frac{5x^3+5x^2+10x^2+10-7x^3-7x^2-7x-7}{5\left(x^2+1\right)\left(x+1\right)}\)

\(=\frac{-2x^3+8x^2-7x+3}{5\left(x^2+1\right)\left(x+1\right)}=\frac{\left(3-x\right)\left(2x^2-2x+1\right)}{5\left(x^2+1\right)\left(x+1\right)}\ge0\)

=> \(f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)(3)

Từ (1); (3) => \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

"=" xảy ra <=> a=3; b=1/3; c=1 và các hoán vị

Trừ vế theo vế hai phương trình trên ta có phương trình:

\(y^2-x^2=x^3-y^3-4x^2+4y^2+3x-3y\)

\(\Leftrightarrow\left(x^3-y^3\right)-3\left(x^2-y^2\right)+\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-3x-3y+3\right)=0\)(1)

\(\Leftrightarrow\orbr{\begin{cases}x-y=0\\x^2+xy+y^2-3x-3y+3=0\end{cases}}\)

+)Với  \(x-y=0\Leftrightarrow x=y\)

Thế vào 1 trong 2 phương trình  ba đầu:

Ta có: \(x^2=x^3-4x^2+3x\Leftrightarrow x^3-5x^2+3x=0\Leftrightarrow x\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5+\sqrt{13}}{2}hoacx=\frac{5-\sqrt{13}}{2}\end{cases}}\)

=> y tự làm nhé 

+) Với \(x^2+xy+y^2-3x-3y+3=0\)

Ta có: \(x^2+xy+y^2-3x-3y+3=\left(x^2+2.x.\frac{y}{2}+\frac{y^2}{4}\right)-3\left(x+\frac{y}{2}\right)+\frac{3y^2}{4}-\frac{3y}{2}+3\)

\(=\left(x+\frac{y}{2}\right)^2-2.\left(x+\frac{y}{2}\right).\frac{3}{2}+\frac{9}{4}+3\left(\frac{y^2}{4}-2.\frac{y}{2}.\frac{1}{2}+\frac{1}{4}\right)-\frac{9}{4}-\frac{3}{4}+3\)

\(=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y}{2}-\frac{1}{2}\right)^2\ge0\)

"=" xảy ra khi và chỉ khi : \(\hept{\begin{cases}x+\frac{y}{2}-\frac{3}{2}=0\\\frac{y}{2}-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)

Thế vào 1 trong hai phương trình ban đầu thấy ko thỏa mãn : 1^2=1^3-4.1^2+3.1 vô lí

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