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\(VT=\left(a+\frac{1}{9b}+\frac{1}{9b}+...+\frac{1}{9b}\right)\left(b+\frac{1}{9c}+\frac{1}{9c}+...+\frac{1}{9c}\right)\left(c+\frac{1}{9a}+\frac{1}{9a}+...+\frac{1}{9a}\right)\)
Lưu ý: Đã tách các số \(\frac{1}{b};\frac{1}{c};\frac{1}{a}\)trong ngoặc thành 9 số hạng bằng nhau
Áp dụng AM-GM:
\(VT\ge10\sqrt[10]{a\left(\frac{1}{9b}\right)^9}.10\sqrt[10]{b\left(\frac{1}{9c}\right)^9}.10\sqrt[10]{c\left(\frac{1}{9a}\right)^9}\)
\(=10^3\sqrt[10]{abc\left(\frac{1}{9a}.\frac{1}{9b}.\frac{1}{9c}\right)^9}\)\(=10^3\sqrt[10]{\frac{abc}{\left(9^3\right)^9.\left(abc\right)^9}}\)\(=10^3\sqrt[10]{\frac{1}{9^{27}.a^8b^8c^8}}\)
\(=\frac{10^3}{\sqrt[10]{9^{27}.a^8b^8c^8}}\)\(=\frac{10^3}{\sqrt[10]{9^{15}.\left(3a\right)^8\left(3b\right)^8\left(3c\right)^8}}=\frac{10^3}{3^3\sqrt[10]{\left(3a.3b.3c\right)^8}}\)
\(\ge\frac{10^3}{3^3\sqrt[10]{\left(\frac{3a+3b+3c}{3}\right)^8}}=\frac{10^3}{3^3\sqrt[10]{\left(\frac{3\left(a+b+c\right)}{3}\right)^8}}=\frac{10^3}{3^3}\left(đpcm\right)\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)
\(A=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)=abc+\frac{1}{abc}+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
BĐT Cauchy cho 3 số dương: \(a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow1\ge3\sqrt[3]{abc}\Leftrightarrow abc\le\frac{1}{27}\Leftrightarrow\frac{1}{abc}\ge27\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.\frac{3}{\sqrt[3]{abc}}=9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
BĐT Cauchy cho 2 số dương: \(abc+\frac{1}{729abc}\ge2\sqrt{abc.\frac{1}{27^2abc}}=\frac{2}{27}\)
Biến đổi A thêm 1 tí nữa: \(A=\left(abc+\frac{1}{729abc}\right)+\frac{728}{729}.\frac{1}{abc}+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+1\)
Thế toàn bộ các BĐT vừa tìm được ở trên vào A:
\(A\ge\frac{2}{27}+\frac{728}{729}.27+9+1=\frac{1000}{27}=\left(\frac{10}{3}\right)^2\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=z\\\sqrt{b^2+c^2}=x\\\sqrt{c^2+a^2}=y\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{y^2+z^2-x^2}{2}\\b=\frac{x^2+z^2-y^2}{2}\\c=\frac{x^2+y^2-z^2}{2}\end{cases}}\)\(\forall\hept{\begin{cases}x,y,z>0\\x+y+z=\sqrt{2017}\end{cases}}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(b+c\le\sqrt{2\left(b^2+c^2\right)}=2x\Rightarrow\frac{a^2}{b+c}\ge\frac{y^2+z^2-x^2}{2\sqrt{2}x}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(2\sqrt{2}\cdot VT\ge\frac{y^2+z^2-x^2}{x}+\frac{y^2+x^2-z^2}{z}+\frac{x^2+z^2-y^2}{y}\)
\(=\frac{y^2}{x}+\frac{z^2}{x}+\frac{y^2}{z}+\frac{x^2}{z}+\frac{x^2}{y}+\frac{z^2}{y}-\left(x+y+z\right)\)
\(\ge\frac{\left(2\left(x+y+z\right)\right)^2}{2\left(x+y+z\right)}-\sqrt{2017}=\sqrt{2017}\)
\(\Rightarrow2\sqrt{2}\cdot VT\ge\sqrt{2017}\Rightarrow VT\ge\frac{\sqrt{2017}}{2\sqrt{2}}=VP\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017.\frac{1}{10}\)
\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)
Mà \(2017=a+b+c\) nên :
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{b}{a+c}\right)=201,7\)
\(3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7\)
\(\Leftrightarrow M=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7-3\)
\(\Leftrightarrow M=198,7\)
Vậy ...
198,7 do ban a