Question

Cho abc là độ dài 3 cạnh tam giác 

cm \(A=\frac{a^3}{b+c-a}+\frac{b^3}{a+c-b}+\frac{c^3}{a+b-c}\ge a^2+b^2+c^2\)

Answer 1

hãy sữ dụng d.lí bất đẳng thức tam giác

Answer 2

\(A=\frac{a^3}{b+c-a}+\frac{b^3}{c+a-b}+\frac{c^3}{a+b-c}=\frac{a^4}{ab+ac-a^2}+\frac{b^4}{ba+bc-b^2}+\frac{c^4}{ca+cb-c^2}\)

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

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

Answer 3

thanks

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