We have: $\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}=\frac{a\sqrt{ac}}{b(2\sqrt{ba}+3c)}=\frac{(\sqrt{\frac{a}{b}})^2}{2\sqrt{\frac{b}{c}}+3\sqrt{\frac{a}{b}}}$Establish similar expressions: ................................................Set: $(\sqrt{\frac{a}{b}};....;.....)=(x;y;z)\Rightarrow xyz=1$By inequality Cauchy: $\frac{x^2}{2y+3z}+\frac{2y+3z}{25}\geq \frac{2x}{5}$Similar; ...................................... $\rightarrow P\geq \frac{1}{5}(x+y+z)\geq \frac{3}{5}$ $\rightarrow P_{min}=\frac{3}{5}$ at $a=b=c

We have: $\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}=\frac{a\sqrt{ac}}{b(2\sqrt{ba}+3c)}=\frac{(\sqrt{\frac{a}{b}})^2}{2\sqrt{\frac{b}{c}}+3\sqrt{\frac{a}{b}}}$Establish similar expressions: ................................................Set: $(\sqrt{\frac{a}{b}};....;.....)=(x;y;z)\Rightarrow xyz=1$By inequality Cauchy: $\frac{x^2}{2y+3z}+\frac{2y+3z}{25}\geq \frac{2x}{5}$Similar; ...................................... $\rightarrow P\geq \frac{1}{5}(x+y+z)\geq \frac{3}{5}$ $\rightarrow P_{min}=\frac{3}{5}$ at $a=b=c./$

Ta có BĐT lđ sau:$\Pi (1+a)\geq (1+\sqrt[3]{abc})^3$ ( cm = biến đổi tương đương)?Ta có: $(ab+bc+ca)^2\geq 3abc(a+b+c)=9abc$$\rightarrow ab+bc+ca\geq 3\sqrt{abc}$$\rightarrow P\leq \frac{2}{3(1+\sqrt{abc})}+\sqrt[3]{\frac{abc}{(1+\sqrt[3]{abc})^3}}$Đặt $\sqrt[6]{abc}=t\rightarrow t\in (0;1]$Xét hàm số:$f(t)=\frac{2}{3(1+t^3)}+\frac{t^2}{1+t^2}$ với $t\in (0;1]$có $f'(t)=..............\geq 0$suy ra hàm số đồng biến trên $(0;1]$$\rightarrow f(t)\leq f(1)=.......$$\rightarrow ...............$

Ta có BĐT lđ sau:$\Pi (1+a)\geq (1+\sqrt[3]{abc})^3$ ( cm = biến đổi tương đương)hoặc: http://diendantoanhoc.net/topic/124445-cmr-1-a1-b1-c-geqslant1sqrt3abc3/Ta có: $(ab+bc+ca)^2\geq 3abc(a+b+c)=9abc$$\rightarrow ab+bc+ca\geq 3\sqrt{abc}$$\rightarrow P\leq \frac{2}{3(1+\sqrt{abc})}+\sqrt[3]{\frac{abc}{(1+\sqrt[3]{abc})^3}}$Đặt $\sqrt[6]{abc}=t\rightarrow t\in (0;1]$Xét hàm số:$f(t)=\frac{2}{3(1+t^3)}+\frac{t^2}{1+t^2}$ với $t\in (0;1]$có $f'(t)=..............\geq 0$suy ra hàm số đồng biến trên $(0;1]$$\rightarrow f(t)\leq f(1)=.......$$\rightarrow ...............$