Ta có: \(B=\sqrt{(x+y)(y+z)(z+x)}\left(\dfrac{\sqrt{y+z}}{x}+\dfrac{\sqrt{z+x}}{y}+\dfrac{\sqrt{x+y}}{z}\right)\)
​\(B=\dfrac{(y+z)\sqrt{(x+y)(x+z)}}{x}+\dfrac{(z+x)\sqrt{(y+z)(y+x)}}{y}+\dfrac{(x+y)\sqrt{(z+x)(z+y)}}{z}\)
Áp dụng BĐT Bunhiacopxky: \((x+y)(x+z)\geq (x+\sqrt{yz})^2\) và tương tự với những biểu thức khác suy ra:
\(B\geq \dfrac{(y+z)(x+\sqrt{yz})}{x}+\dfrac{(z+x)(y+\sqrt{xz})}{y}+\dfrac{(x+y)(z+\sqrt{xy})}{z}\) hay \(B\geq 2(x+y+z)+\dfrac{(y+z)\sqrt{yz}}{x}+\dfrac{(z+x)\sqrt{zx}}{y}+\dfrac{(x+y)\sqrt{xy}}{z}\) hay \(B\geq 2(x+y+z)+\underbrace{\dfrac{yz(y+z)\sqrt{yz}+xz(x+z)\sqrt{xz}+xy(x+y)\sqrt{xy}} {xyz}}_{M}\)
Đặt \((x,y,z)=(a^2,b^2,c^2)\)
Khi đó: \(M=\dfrac{a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(a^2+c^2)}{a^2b^2c^2}\)
Áp dụng BĐT AM-GM: \(a^5b^3+a^3b^5\geq 2\sqrt{a^8b^8}=2a^4b^4\) \(b^5c^3+c^5b^3\geq 2b^4c^4\) \(c^5a^3+a^5c^3\geq 2c^4a^4\) \(\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2(a^4b^4+b^4c^4+c^4a^4)\) (1) (cộng các BĐT theo vế)
Tiếp tục AM-GM: \(a^4b^4+b^4c^4\geq 2a^2b^4c^2; b^4c^4+c^4a^4\geq 2a^2b^2c^4; c^4a^4+a^4b^4\geq 2a^4b^2c^2\) \(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^2c^2(a^2+b^2+c^2)\) (2)
Từ\((1); (2)\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2a^2b^2c^2(a^2+b^2+c^2)\) \(\Rightarrow M\geq 2(a^2+b^2+c^2)=2(x+y+z)\) Do đó: \(A\geq 2(x+y+z)+M\geq 4(x+y+z)\Leftrightarrow B\geq 4\sqrt{2021}\)