Đặt $y=az;z=bx\Rightarrow a;b\in [\frac{1}{4};1].$Khi đó: $P=\frac{1}{2+3a}+\frac{a}{a+b}+\frac{b}{b+1}.$
Xét hàm số: $f(a)=\frac{1}{2+3a}+\frac{a}{a+b}; f'(a)=-\frac{3}{(2+3a)^2}+\frac{b}{(a+b)^2}.$
Xét $b(2+3a)^2-3(a+b)^2=9a^2b+6ab+4b-3a^2-3b^2\geq 15a^2b+4b-3a^2-3b^2=3a^2(5b-1)+b(4-3b)>0$
Nên $f(a)$ là hàm đồng biến trên $[\frac{1}{4};1]\Rightarrow f(a)\geq f(\frac{1}{4})=\frac{4}{11}+\frac{1}{1+4b}$
Do đó: $P\geq \frac{4}{11}+\frac{1}{1+4b}+\frac{b}{b+1}=g(b)$
Ta có $g'(b)=\frac{-4}{(1+4b)^2}+\frac{1}{(b+1)^2}\Rightarrow g'(b)=0\rightarrow b=\frac{1}{2}$
Tuqf đó suy ra $g(b)\geq g(\frac{1}{2})=\frac{34}{33}$ hay $P\geq \frac{34}{33}$