\(a+b+c=\dfrac{ab+ac}{2}=\dfrac{a\left(b+c\right)}{2}\Rightarrow\dfrac{b+c}{2}=\dfrac{a+b+c}{a}\)
\(\Rightarrow\dfrac{2}{b+c}=\dfrac{a}{a+b+c}\)
Tương tự ta có: \(\dfrac{3}{c+a}=\dfrac{b}{a+b+c}\)
\(\dfrac{4}{a+b}=\dfrac{c}{a+b+c}\)
Cộng vế với vế:
\(\dfrac{2}{b+c}+\dfrac{3}{c+a}+\dfrac{4}{a+b}=\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
Vậy \(P=1\)
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