1.ta có $a+\sqrt{ab}+\sqrt[3]{abc}=a+\sqrt{\frac{1}{2}a.2b} +\sqrt[3]{\frac{1}{4}a.b.4c}$ $\leq a+\frac{1}{4}a+b+\frac{1}{12}a+ \frac{b}{3}+\frac{4}{3}c =\frac{4}{3}(a+b+c)$$\Rightarrow P\geq \frac{3}{2(a+b+c)}-\frac{3}{\sqrt{a+b+c}}$ Đặt $t=\sqrt{a+b+c}, t>0$P $\geq \frac{3}{2t^{2}}-\frac{3}{t}=\frac{3}{2}(\frac{1}{t}-1)^{2}-\frac{3}{2}\geq \frac{-3}{2}$Dấu '=' $\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}$
1.ta có $a+\sqrt{ab}+\sqrt[3]{abc}=a+\sqrt{\frac{1}{2}a.2b} +\sqrt[3]{\frac{1}{4}a.b.4c}$ $\leq a+\frac{1}{4}a+b+\frac{1}{12}a+ \frac{b}{3}+\frac{4}{3}c =\frac{4}{3}(a+b+c)$$\Rightarrow P\geq \frac{3}{2(a+b+c)}-\frac{3}{\sqrt{a+b+c}}$ Đặt $t=\sqrt{a+b+c}, t>0$P $\geq \frac{3}{2t^{2}}-\frac{3}{t}=\frac{3}{2}(\frac{1}{t}-1)^{2}-\frac{3}{2}\geq \frac{-3}{2}$Dấu '=' $\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}4
1.ta có $a+\sqrt{ab}+\sqrt[3]{abc}=a+\sqrt{\frac{1}{2}a.2b} +\sqrt[3]{\frac{1}{4}a.b.4c}$ $\leq a+\frac{1}{4}a+b+\frac{1}{12}a+ \frac{b}{3}+\frac{4}{3}c =\frac{4}{3}(a+b+c)$$\Rightarrow P\geq \frac{3}{2(a+b+c)}-\frac{3}{\sqrt{a+b+c}}$ Đặt $t=\sqrt{a+b+c}, t>0$P $\geq \frac{3}{2t^{2}}-\frac{3}{t}=\frac{3}{2}(\frac{1}{t}-1)^{2}-\frac{3}{2}\geq \frac{-3}{2}$Dấu '=' $\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}
$