Ta sẽ chứng minh BĐT phụ sau : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{3^2}{a+b+c}=\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Trong đó : \(\hept{\begin{cases}a=x+y\\b=y+z\\c=z+x\end{cases}}\) , Ta có :
\(\left(x+y+y+z+x+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge9\)
\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge4,5\)
\(\Leftrightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}\ge4,5\)
\(\Leftrightarrow1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{x+z}\ge4,5\)
\(\Leftrightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{z+y}\ge1,5\)
\(\Rightarrow P_{min}=1,5\) " = " \(\Leftrightarrow x=y=z\)
Chúc bạn học tốt !!!
