Cách 1Dùng bdt holder ta có $P^2.6=(a^3+2b^3+3c^3)(a^3+2b^3+3c^3)(1+2+3) \ge (a^2+2b^2+3c^2)^3=1$$\Leftrightarrow P \ge \sqrt{\frac 16}$Đẳng thức xảy ra $\Leftrightarrow a=b=c=\sqrt{\frac 16}$Cách 2Dùng bdt AM-GM$\left.\begin{matrix} a^3+a^3+\dfrac 1{\sqrt{6^3}} \ge \dfrac{3}{\sqrt 6}a^2 \\ 2b^3+2b^3+\dfrac{2}{\sqrt{6^3}} \ge \dfrac{6}{\sqrt 6}b^2 \\ 3c^3+3c^3+ \dfrac{3}{\sqrt{6^3}} \ge \dfrac{9}{\sqrt 6}c^2\end{matrix}\right\}\Rightarrow \text{kết quả tương tự}$

Dùng bdt holder ta có $P^2.\frac{407}{144}=(2a^3+3b^3+4c^3)(2a^3+3b^3+4c^3)(\frac 14+\frac 89+\frac{27}{16}) \ge (a^2+2b^2+3c^2)^3=1$$\Leftrightarrow P \ge \frac{12}{\sqrt{407}}$Đẳng thức xảy ra $\Leftrightarrow a=\frac{6}{\sqrt{407}},b=\frac{8}{\sqrt{407}},c=\frac{9}{\sqrt{407}}$

$bdt\Leftrightarrow \sqrt[n]{\frac{a}{(b+c)(n-1)}}+\sqrt[n]{\frac{b}{(c+a)(n-1)}}+\sqrt[n]{\frac{c}{(a+b)(n-1)}} > \frac{n}{n-1}$Ta có :$\sqrt[n]{\frac{a}{(b+c)(n-1)}}=\frac{1}{\sqrt[n]{\left[\dfrac {b+c}a(n-1) \right].\overset{n-1 \hspace{1mm} \text{số} \hspace{1mm}1}{\overbrace{1.1...1}}}} \ge \frac{n}{\dfrac{b+c}a(n-1)+\overset{n-1 \hspace{1mm} \text{số} \hspace{1mm} 1}{\overbrace{1+1+1+...+1}}}$$\frac{an}{(a+b+c)(n-1)}$Tương tự cộng lại $\Rightarrow$ dpcm (dấu = ko xảy ra)

$bdt\Leftrightarrow \sqrt[n]{\frac{a}{(b+c)(n-1)}}+\sqrt[n]{\frac{b}{(c+a)(n-1)}}+\sqrt[n]{\frac{c}{(a+b)(n-1)}} > \frac{n}{n-1}$Ta có :$\sqrt[n]{\frac{a}{(b+c)(n-1)}}=\frac{1}{\sqrt[n]{\left[\dfrac {b+c}a(n-1) \right].\overset{n-1 \hspace{1mm} \text{số} \hspace{1mm}1}{\overbrace{1.1...1}}}} \ge \frac{n}{\dfrac{b+c}a(n-1)+\overset{n-1 \hspace{1mm} \text{số} \hspace{1mm} 1}{\overbrace{1+1+1+...+1}}}$$=\frac{an}{(a+b+c)(n-1)}$Tương tự cộng lại $\Rightarrow$ dpcm (dấu = ko xảy ra)Hoặc xem tại đây