BĐT cần chứng minh tương đương
\(VT\ge4\left(x+y+z\right)\)
\(\Leftrightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Theo BĐT Cauchy-Schwarz và AM-GM, ta có:
\(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge\dfrac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}=y+z+\dfrac{\left(y+z\right)\sqrt{yz}}{x}\ge y+z+\dfrac{2yz}{x}\)
Suy ra: \(\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge2\left(x+y+z\right)-2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\)
Mặt khác, theo AM-GM:
\(\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)^2\ge3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge x+y+z\)
\(\Rightarrow\sum\dfrac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge4\left(x+y+z\right)\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{\sqrt{2}}{3}\)
@Phương An