Question

cho a,b,c là 3 cạnh tam giác

chứng minh

\(\frac{a}{2\left(b+c\right)-a}+\frac{b}{2\left(a+c\right)-b}+\frac{c}{2\left(a+b\right)-c}\ge1\)

Answer 1

\(P=\frac{a}{2\left(b+c\right)-a}+\frac{b}{2\left(c+a\right)-b}+\frac{c}{2\left(a+b\right)-c}\)

\(=\frac{a^2}{2\left(ab+ca\right)-a^2}+\frac{b^2}{2\left(bc+ab\right)-b^2}+\frac{c^2}{2\left(ca+bc\right)-c^2}\)

\(\ge\frac{\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{\frac{4}{3}\left(a+b+c\right)^2-\frac{\left(a+b+c\right)^2}{3}}=1\)

      

Answer 2

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)......................................................

Answer 3

\(P=\frac{a}{2\left(b+c\right)-a}+\frac{b}{2\left(c+a\right)-b}+\frac{c}{2\left(a+b\right)-c}\)

\(=\frac{a^2}{2\left(ab+ca\right)-a^2}+\frac{b^2}{2\left(bc+ab\right)-b^2}+\frac{c^2}{2\left(ca+bc\right)-c^2}\)

\(\ge\frac{\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{\frac{4}{3}\left(a+b+c\right)^2-\frac{\left(a+b+c\right)^2}{3}=1}\)

Chúc bạn học tốt!     @_@