Lời giải:
Vì tam giác có chu vi bằng $1$ nên $a+b+c=1$
\(\Rightarrow 1-a, 1-b, 1-c>0\)
Thay vào biểu thức đã cho:
\(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{3}{2}\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\right)[a(1-a)+b(1-b)+c(1-c)]\geq (a+b+c)^2\)
\(\Leftrightarrow \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{(a+b+c)^2}{(a+b+c)-(a^2+b^2+c^2)}\)
\(\Leftrightarrow \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1}{1-(a^2+b^2+c^2)}\)
Áp dụng BĐT Bunhiacopxky: \((a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2=1\)
\(\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}\)
\(\Rightarrow 1-(a^2+b^2+c^2)\leq \frac{2}{3}\)
Suy ra \( \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1}{1-(a^2+b^2+c^2)}\geq \frac{1}{\frac{2}{3}}=\frac{3}{2}\)
Dấu bằng xảy ra khi \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\Leftrightarrow a=b=c\) hay tam giác $ABC$ đều.
Cách khác:v
Giải: \(gt:\left\{{}\begin{matrix}a;b;c>0\\a+b+c=1\end{matrix}\right.\)
\(pt\Leftrightarrow\dfrac{a}{\left(a+b+c\right)-a}+\dfrac{b}{\left(a+b+c\right)-b}+\dfrac{c}{\left(a+b+c\right)-c}=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{3}{2}\)
\(Nesbit:\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Dấu "=" \(a=b=c\Leftrightarrow\Delta ABC\) đều
