Ta sẽ cm bđt:\(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Áp dụng bđt bunhia:
\(\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\right)\left(a+b+c\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Áp dụng vào suy ra:
\(A=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{2}{2}=1\)
"="<=>x=y=z=\(\dfrac{2}{3}\)
Cách khác:
Áp dụng BĐT Cauchy cho các số dương ta có:
\(\frac{x^2}{y+z}+\frac{y+z}{2}\geq 2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}=y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}=z\)
Cộng theo vế và rút gọn ta có:
\(A+\frac{x+y+z}{2}\geq x+y+z\)
\(\Rightarrow A\geq \frac{x+y+z}{2}=1\)
Vậy \(A_{\min}=1\Leftrightarrow x=y=z=\frac{2}{3}\)
