Ta có:
\(\frac{xz}{y^2+yz}+\frac{y^2}{xz+yz}+\frac{x+2z}{x+z}=\frac{xz}{y^2+yz}+\frac{y^2}{xz+yz}+\frac{z}{x+z}+1\)
\(=\frac{1}{\frac{y^2}{xz}+\frac{y}{x}}+\frac{1}{\frac{xz}{y^2}+\frac{z}{y}}+\frac{1}{\frac{x}{z}+1}+1\)
Đặt \((\frac{x}{y}, \frac{y}{z})=(a,b)\Rightarrow ab=\frac{x}{z}\geq 1\) do $x\ge z$
Bài toán trở thành: Cho 2 số dương $a,b$ thỏa mãn $ab\geq 1$. Tìm min của
\(P=\frac{1}{\frac{b}{a}+\frac{1}{a}}+\frac{1}{\frac{a}{b}+\frac{1}{b}}+\frac{1}{ab+1}+1=\frac{a}{b+1}+\frac{b}{a+1}+\frac{1}{ab+1}+1\)
Có: \(P+1=\frac{a+b+1}{b+1}+\frac{b+a+1}{a+1}+\frac{1}{ab+1}\). Áp dụng BĐT Cauchy-Schwarz và AM-GM ta có:
\(P+1\geq (a+b+1).\frac{4}{b+1+a+1}+\frac{1}{(\frac{a+b}{2})^2+1}=\frac{4(a+b+1)}{a+b+2}+\frac{4}{(a+b)^2+4}(1)\)
Đặt \(t=a+b\). Theo BĐT AM-GM \(t=a+b\geq 2\sqrt{ab}\geq 2\sqrt{1}=2\)
Xét hiệu:
\(\frac{4(a+b+1)}{a+b+2}+\frac{4}{(a+b)^2+4}-\frac{7}{2}=\frac{4(t+1)}{t+2}+\frac{4}{t^2+4}-\frac{7}{2}\)
\(=\frac{t^3-6t^2+12t-8}{2(t+2)(t^2+4)}=\frac{(t-2)^3}{2(t+2)(t^2+4)}\geq 0, \forall t\geq 2\)
\(\Rightarrow \frac{4(a+b+1)}{a+b+2}+\frac{4}{(a+b)^2+4}\geq \frac{7}{2}(2)\)
Từ \((1);(2)\Rightarrow P+1\geq \frac{7}{2}\Rightarrow P\geq \frac{5}{2}\)
Vậy $P_{\min}=\frac{5}{2}$
Dấu "=" xảy ra khi $x=y=z$
