\(\left(4+\dfrac{1}{4}\right)\left(a^2+\dfrac{1}{b+c}\right)\ge\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)^2\)
\(\Rightarrow\sqrt{a^2+\dfrac{1}{b+c}}\ge\dfrac{2}{\sqrt{17}}\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)=\dfrac{1}{\sqrt{17}}\left(4a+\dfrac{1}{\sqrt{b+c}}\right)\)
Tương tự:
\(\sqrt{b^2+\dfrac{1}{a+c}}\ge\dfrac{1}{\sqrt{17}}\left(4b+\dfrac{1}{\sqrt{a+c}}\right)\) ; \(\sqrt{c^2+\dfrac{1}{a+b}}\ge\dfrac{1}{\sqrt{17}}\left(4c+\dfrac{1}{\sqrt{a+b}}\right)\)
Cộng vế:
\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\right)\)
\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)\)
Cũng theo Bunhiacopxki:
\(1.\sqrt{a+b}+1.\sqrt{b+c}+1\sqrt{c+a}\le\sqrt{\left(1+1+1\right)\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\)
\(\Rightarrow VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{6\left(a+b+c\right)}}\right)\)
\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}\left(a+b+c\right)+\dfrac{a+b+c}{8}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}\right)\)
\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}.6+3\sqrt[3]{\dfrac{81\left(a+b+c\right)}{32.6\left(a+b+c\right)}}\right)=\dfrac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=2\)