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3) Ta  có \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}\)

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

Ta dễ chứng minh được rằng \(\frac{a}{c}+\frac{c}{a}\ge2\)

Thật vậy \(\frac{a}{c}+\frac{c}{a}\ge2\)

<=> \(\frac{a^2+c^2}{ac}\ge2\)

<=> a² + c² \(\ge\)2ac

<=> (a - c)² \(\ge0\)(đúng với a,c > 0)

Tương tự \(\hept{\begin{cases}\frac{b}{a}+\frac{a}{b}\ge2\\\frac{b}{c}+\frac{c}{b}\ge2\end{cases}}\)

Khi đó \(\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)\ge2+2+2=6\)(đpcm)

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Đặt \(x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}, \Rightarrow x+y+z=2\)

Suy ra    \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\)

Ta có \(\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{\left(2-x\right)^2} .\frac{2-x}{8}.\frac{2-x}{8}}=\frac{3x}{4}.\)

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

dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)hay \(a=b=c=\frac{3}{2}\)