Đặt \(\left(a+b-c;a-b+c;-a+b+c\right)=\left(x;y;z\right)\)
\(\Rightarrow p=\frac{a+b+c}{2}=\frac{x+y+z}{2}\) ; \(p-a=\frac{-a+b+c}{2}=\frac{z}{2}\) ...
\(\Rightarrow S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\sqrt{\frac{xyz\left(x+y+z\right)}{16}}=\frac{1}{4}\sqrt{xyz\left(x+y+z\right)}\)
\(\Rightarrow\sqrt{S}=\frac{1}{2}\sqrt[4]{xyz\left(x+y+z\right)}\)
BĐT trở thành:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[4]{\frac{3}{xyz\left(x+y+z\right)}}\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\)
Nên chỉ cần chứng minh: \(\frac{1}{\sqrt[3]{xyz}}\ge\sqrt[4]{\frac{3}{xyz\left(x+y+z\right)}}\)
Mũ 12 hai vế: \(\Leftrightarrow\frac{1}{\left(xyz\right)^4}\ge\frac{27}{\left(xyz\right)^3\left(x+y+z\right)^3}\Leftrightarrow\left(x+y+z\right)^3\ge27xyz\)
Hiển nhiên đúng theo AM-GM
Dấu "=" xảy ra khi tam giác đều
