Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{y+z}{2}\\b=\frac{x+z}{2}\\c=\frac{x+y}{2}\end{matrix}\right.\)
a) BĐT (1) \(\Leftrightarrow xyz\le\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8}\)
BĐT này luôn đúng theo Cô-si :
\(VP\ge\frac{2\sqrt{xy}\cdot2\sqrt{yz}\cdot2\sqrt{xz}}{8}=\frac{8xyz}{8}=xyz\)
Dấu "=" khi tam giác ABC đều
b) BĐT (2) \(\Leftrightarrow\frac{\frac{x+y}{2}}{z}+\frac{\frac{y+z}{2}}{x}+\frac{\frac{x+z}{2}}{y}\ge3\)
\(\Leftrightarrow\frac{x+y}{2z}+\frac{y+z}{2x}+\frac{x+z}{2y}\ge3\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{x+y}{z}+\frac{y+z}{x}+\frac{x+z}{y}\right)\ge3\)
\(\Leftrightarrow\frac{x+y}{z}+\frac{y+z}{x}+\frac{x+z}{y}\ge6\)
Áp dụng bđt Cô-si 3 số :
\(VT\ge3\sqrt[3]{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{xyz}}=3\sqrt[3]{\frac{8xyz}{xyz}}=3\cdot\sqrt[3]{8}=3\cdot2=6\)( theo kết quả câu a )
=> đpcm
Dấu "=" khi tam giác ABC đều
c) Áp dụng bđt Cauchy-Schwarz:
\(VT\ge\frac{\left(a+b+c\right)^2}{a+b-c+b+c-a+a+c-b}=\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" khi tam giác ABC đều
