Điểm rơi: $a=c=\dfrac{1}{7}b$(Điểm rơi này khá khoai, cần nhiều kĩ thuật để xác định)

Ta tách và Cauchy như sau:

`P=\sqrt{a^2/((a+b)(a+c))}+\sqrt{(4b^2)/((b+c)(a+b))}+\sqrt{c^2/((c+a)(c+b))}`

`=2\sqrt{a/(a+b) . a/(4(a+c))}+\sqrt{(2b)/(b+c) . (2b)/(a+b)} + 2\sqrt{c/(4(c+a)) . c/(b+c)}`

Áp dụng BĐT Cauchy ta được:

`P\le a/(a+b)+a/(4(a+c)) + b/(b+c)+b/(a+b) + c/(4(c+a))+c/(b+c)`

`\le (a/(a+b)+b/(a+b)) + (a/(4(a+c)) + c/(4(a+c))) + (b/(b+c)+c/(b+c))`

`\le 1+1/4+1`

`\le 9/4`

Dấu "=" xảy ra khi: `a=c=1/7b`

Theo Cauchy:

\(\begin{array}{l}
P=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{2b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\\
\to P=\sqrt{\dfrac{a}{2\left(a+c\right)}}\cdot \sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}}\cdot \sqrt{\dfrac{2b}{b+a}}+\sqrt{\dfrac{c}{2\left(a+c\right)}}\cdot \sqrt{\dfrac{2c}{c+b}}\\
\to P\le \dfrac{1}{2}\left[\dfrac{a}{2\left(a+c\right)}+\dfrac{2a}{a+b}+\dfrac{2b}{b+c}+\dfrac{2b}{b+a}+\dfrac{a}{2\left(a+c\right)}+\dfrac{2c}{c+b}\right]\\
\to P\le \dfrac{1}{2}\left[\dfrac{a+c}{2\left(a+c\right)}+\dfrac{2\left(a+b\right)}{a+b}+\dfrac{2\left(b+c\right)}{b+c}\right]\\
\to P\le \dfrac{1}{2}\left(\dfrac{1}{2}+2+2\right)=\dfrac{9}{4}
\end{array}\)

Dấu \("="\leftrightarrow \begin{cases} c^2+bc=4c^2+4ac\\ 2b^2+2ab=2b^2+2bc\\ 4a^2+4ac=a^2+ab \end{cases}\leftrightarrow a=c=\dfrac{b}{7}\)