Ta có \(\sqrt{1+\dfrac{16a}{b+c}}=\sqrt{1+\dfrac{16a^2}{ab+ac}}\)\(\)
\(\Rightarrow3.\sqrt{1^2+\left(\dfrac{4a}{\sqrt{ab+ac}}\right)^2}=\sqrt{\left(1^2+\left(\dfrac{4a}{\sqrt{ab+ac}}\right)\right).\left(1^2+\left(2\sqrt{2}\right)^2\right)}\ge1+\dfrac{8a\sqrt{2}}{\sqrt{ab+ac}}\)Dấu "=" xảy ra <=> \(\dfrac{\dfrac{4a}{\sqrt{ab+ac}}}{2\sqrt{2}}=1\Leftrightarrow\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{2}}{2}\Leftrightarrow\dfrac{a}{b+c}=\dfrac{1}{2}\)
Tương tự ta có \(3.\sqrt{1+\dfrac{16a}{b+c}}+3.\sqrt{1+\dfrac{16b}{a+c}}+3.\sqrt{1+\dfrac{16c}{a+b}}\)
\(\ge3+\dfrac{8a\sqrt{2}}{\sqrt{ab+ac}}+\dfrac{8b\sqrt{2}}{\sqrt{ab+bc}}+\dfrac{8c\sqrt{2}}{\sqrt{ac+bc}}\)
\(=3+8\sqrt{2}.\left(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\right)\)
Xét \(A=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)
Ta có \(\sqrt{\dfrac{a}{b+c}}=\dfrac{a\sqrt{2}}{\sqrt{2a}.\sqrt{b+c}}\ge\dfrac{a\sqrt{2}}{\dfrac{2a+b+c}{2}}=\dfrac{2\sqrt{2}.a}{2a+b+c}\)(BĐT Cauchy)
Tương tự \(A\ge\dfrac{2\sqrt{2}.a}{2a+b+c}+\dfrac{2\sqrt{2}.b}{2b+a+c}+\dfrac{2\sqrt{2}.c}{2c+a+b}\)
\(=\sqrt{2}.\left(\dfrac{2a}{2a+b+c}+\dfrac{2b}{2b+a+c}+\dfrac{2c}{2c+a+b}\right)\)
\(=\sqrt{2}\left(3-\dfrac{b+c}{2a}\right)\)
