Question

Cho a;b;c là 3 cạnh tam giác. Chứng minh:

\(\sqrt{\dfrac{a}{b+c-a}}+\sqrt{\dfrac{b}{c+a-b}}+\sqrt{\dfrac{c}{a+b-c}}\ge3\)

Answer 1

Áp dụng BĐT AM-GM ta có:

\(2\sqrt{\dfrac{y+z-x}{x}}\le\dfrac{y+z-x}{x}+1=\dfrac{y+z}{x}\)

\(\Leftrightarrow\sqrt{\dfrac{x}{y+z-x}}\ge\dfrac{2x}{y+z}\)

Áp dụng vào đề bài ta có:

\(A=\sqrt{\dfrac{a}{b+c-a}}+\sqrt{\dfrac{b}{c+a-b}}+\sqrt{\dfrac{c}{a+b-c}}\ge\)

\(\ge\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=\dfrac{2.3}{2}=3\)(BĐT Nesbitt)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)