\(a^2\le bc\le\frac{\left(b+c\right)^2}{4}\Rightarrow\frac{b+c}{a}\ge2\)
\(P=\frac{1}{a}\left(\frac{b^2}{c}+\frac{c^2}{b}\right)+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}}\)
\(P\ge\frac{\left(b+c\right)^2}{a\left(b+c\right)}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{\left(b+c\right)^2}{a\left(b+c\right)+2bc}}\)
\(P\ge\frac{b+c}{a}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{\left(b+c\right)^2}{a\left(b+c\right)+\frac{\left(b+c\right)^2}{2}}}\)
\(P\ge\frac{b+c}{a}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{2}{\frac{2a}{b+c}+1}}\)
Đặt \(\frac{b+c}{a}=x\ge2\)
\(\Rightarrow P\ge x+\frac{1}{\sqrt{x}}+\sqrt{\frac{2}{\frac{2}{x}+1}}=\left(1-\frac{1}{4\sqrt{2}}\right)x+\frac{x}{4\sqrt{2}}+\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}+\sqrt{\frac{2}{\frac{2}{x}+1}}\)
\(P\ge\left(1-\frac{1}{4\sqrt{2}}\right).2+3\sqrt[3]{\frac{x}{16x\sqrt{2}}}+\sqrt{\frac{2}{\frac{2}{2}+1}}\)
\(P\ge2-\frac{1}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}+1=3+\frac{1}{2\sqrt{2}}\)
Dấu "=" xảy ra khi \(a=b=c\)
