Ta có \(2\sqrt{(x^{4}+y+z)(xy+yz+zx)}=2\sqrt{(x^{4}+xy^{2}z+xyz^{2})(xy+yz+zx)}\)
\(=2\sqrt{(x^{3}+y^{2}z+yz^{2})(x^{2}y+zx^{2}+1)}\leq x^{3}+1+(y+z)(\frac{1}{x}+x^{2})\)
\(=(x^{3}+1)\frac{x+y+z}{x}\)
Chứng minh tương tự
\(\frac{x^{3}+1}{\sqrt{x^{4}+y+z}}\geq 2\sqrt{xy+yz+zx}.\frac{x}{x+y+z}\)
\(\frac{y^{3}+1}{\sqrt{y^{4}+z+x}}\geq 2\sqrt{xy+yz+zx}.\frac{y}{x+y+z}\)
\(\frac{z^{3}+1}{\sqrt{z^{4}+x+y}}\geq 2\sqrt{xy+yz+zx}.\frac{z}{x+y+z}\)
Do đó
\(P\geq 2\sqrt{xy+yz+zx}-\frac{8(xy+yz+zx)}{xy+yz+zx+1}\)
Đặt \(xy+yz+zx=t(t\geq 3)\; (Do\; xy+yz+zx\geq 3\sqrt{x^{2}y^{2}z^{2}}=3)\)
Suy ra \(P\geq 2\sqrt{t}-\frac{8t}{t+1}.\) Xét hàm số
\(f(t)=2\sqrt{t}-\frac{8t}{t+1};t\in [3;+\infty )\Rightarrow f't=\frac{1}{\sqrt{t}}-\frac{8}{(t+1)^{2}}=\frac{t^{2}+2t+1-8\sqrt{t}}{\sqrt{t}(t+1)^{2}}\)
Lại có: \(t^{2}+2t+1-8\sqrt{t}\geq 5t+1-8\sqrt{t}> 5\sqrt{3}\sqrt{t}-8\sqrt{t}> 0,\forall t\in [3;+\infty )\)
\(\Rightarrow f'(t)> 0,\forall t\in [3;+\infty ),f(t)\) liên tục trên \([3;+\infty )\)
\(\Rightarrow\) Hàm số \(f(t)\) đồng biến trên \([3;+\infty )\)
\(P\geq f(3)=2\sqrt{3}-6\)
Vậy \(minP=2\sqrt{3}-6\) đạt được khi \(x=y=z=1\)
