Question

Tam giác ABC có \(\frac{a^3+b^3+c^3}{abc}+\frac{2r}{R}=4\) CM ABC là tam giác đều

Answer 1

Chu y: \(\frac{a^3+b^3+c^3}{abc}+\frac{2r}{R}=\frac{a^3+b^3+c^3}{abc}+\frac{\frac{2S}{\frac{a+b+c}{2}}}{\frac{abc}{4S}}\)

\(=\frac{16S^2}{abc\left(a+b+c\right)}+\frac{a^3+b^3+c^3}{abc}=\frac{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}{abc}+\frac{a^3+b^3+c^3}{abc}\)

Khi dok \(\frac{a^3+b^3+c^3}{abc}+\frac{2r}{R}-4=\frac{a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-6abc}{abc}\ge\frac{6abc-6abc}{abc}\ge0\)

\(\Rightarrow\frac{a^3+b^3+c^3}{abc}+\frac{2r}{R}\ge4\)

"=" khi \(a=b=c\leftrightarrow\Delta ABC \text{đều}\)