2 BĐT này đều áp dụng Mincopxki là ra thôi:
\(A=\sqrt{a^2+\dfrac{1}{b+c}}+\sqrt{b^2+\dfrac{1}{c+a}}+\sqrt{c^2+\dfrac{1}{a+b}}\)
\(A\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}+\dfrac{1}{\sqrt{a+b}}\right)^2}\)
Đặt \(P=\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\ge\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\)
\(P\ge\dfrac{9}{\sqrt{3\left(a+b+b+c+c+a\right)}}=\dfrac{9}{\sqrt[]{6\left(a+b+c\right)}}\)
\(\Rightarrow A\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{6\left(a+b+c\right)}}\)
\(\Rightarrow A\ge\sqrt{\dfrac{31\left(a+b+c\right)^2}{32}+\dfrac{1}{32}\left(a+b+c\right)^2+\dfrac{81}{12\left(a+b+c\right)}+\dfrac{81}{12\left(a+b+c\right)}}\)
\(\Rightarrow A\ge\sqrt{\dfrac{31}{32}.6^2+3\sqrt[3]{\dfrac{81^2\left(a+b+c\right)^2}{32.12^2}}}=\dfrac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=2\)
\(B=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(B\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(B\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}\)
\(B\ge\sqrt{\dfrac{15}{16}\left(a+b+c\right)^2+\dfrac{1}{16}\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(B\ge\sqrt{\dfrac{15}{16}.6^2+2\sqrt{\dfrac{81\left(a+b+c\right)^2}{16\left(a+b+c\right)^2}}}=\dfrac{3\sqrt{17}}{2}\)
