\(VT\le\frac{1}{2\sqrt{a^2bc}}+\frac{1}{2\sqrt{b^2ac}}+\frac{1}{2\sqrt{c^2ab}}=\frac{1}{2}\left(\frac{1}{\sqrt{ab.ac}}+\frac{1}{\sqrt{ab.bc}}+\frac{1}{\sqrt{ac.bc}}\right)\)
\(VT\le\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{bc}\right)=\frac{1}{2}\left(\frac{a+b+c}{abc}\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
Tất cả đều là BĐT Cô-si đó bạn:
\(a^2+bc\ge2\sqrt{a^2bc}\Rightarrow\frac{1}{a^2+bc}\le\frac{1}{2\sqrt{a^2bc}}\)
\(\frac{1}{\sqrt{ab.ac}}=\sqrt{\frac{1}{ab}}.\sqrt{\frac{1}{ac}}\le\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{ac}\right)\) (chính là BĐT Cô-si dạng \(\sqrt{xy}\le\frac{1}{2}\left(x+y\right)\) thôi)
