\(GT\Rightarrow a^2=b^2+c^2\ge\frac{1}{2}\left(b+c\right)^2\Rightarrow\frac{a}{b+c}\ge\frac{\sqrt{2}}{2}\)
\(M=8a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{b+c}{a}+2019\ge4a^2\left(\frac{1}{b}+\frac{1}{c}\right)^2+\frac{b+c}{a}+2019\)
\(M\ge4a^2\left(\frac{4}{b+c}\right)^2+\frac{b+c}{a}+2019=16\left(\frac{a}{b+c}\right)^2+\frac{b+c}{a}+2019\)
Đặt \(\frac{a}{b+c}=x\ge\frac{\sqrt{2}}{2}\)
\(\Rightarrow M\ge16x^2+\frac{1}{x}+2019=\sqrt{2}x^2+\frac{1}{2x}+\frac{1}{2x}+\left(16-\sqrt{2}\right)x^2+2019\)
\(\Rightarrow M\ge3\sqrt[3]{\frac{\sqrt{2}x^2}{4x^2}}+\left(16-\sqrt{2}\right).\frac{1}{2}+2019=2027+\sqrt{2}\)
Dấu "=" xảy ra khi \(b=c=\frac{a}{\sqrt{2}}\)
