1,$A=a+b+c-(\frac{2ab^{3}}{a+2b^{3}}+\frac{2bc^{3}}{b+2c^{3}}+\frac{2ca^{3}}{c+2a^{3}})$$a+2b^{3}=a+b^{3}+b^{3}\geq 3b^{2}\sqrt[3]{a}\rightarrow \frac{2ab^{3}}{a+2b^{3}}\leq \frac{2}{3}.\sqrt[3]{ab.ab.b}\leq \frac{2}{9}(2ab+b)$tương tự cuối cùng được: $A\geq a+b+c-\frac{2}{9}(2ab+2bc+2ca+a+b+c)\geq 3-\frac{2}{9}.(2.\frac{9}{3}+3)=1$2,$A=\frac{a^{4}}{a^{2}+2bca^{2}}+\frac{b^{4}}{b^{2}+2acb^{2}}+\frac{c^{4}}{c^{2}+2abc^{2}}\geq \frac{1}{1+2abc(a+b+c)}\geq \frac{1}{1+\frac{2.9(a^{2}+b^{2}+c^{2})^{2}}{27}}\geq \frac{3}{5}$
1,$A=a+b+c-(\frac{2ab^{3}}{a+2b^{3}}+\frac{2bc^{3}}{b+2c^{3}}+\frac{2ca^{3}}{c+2a^{3}})$$a+2b^{3}=a+b^{3}+b^{3}\geq 2b^{2}\sqrt[3]{a}\rightarrow \frac{2ab^{3}}{a+2b^{3}}\leq \frac{2}{3}.\sqrt[3]{ab.ab.b}\leq \frac{2}{9}(2ab+b)$tương tự cuối cùng được: $A\geq a+b+c-\frac{2}{9}(2ab+2bc+2ca+a+b+c)\geq 3-\frac{2}{9}.(2.\frac{9}{3}+3)=1$2,$A=\frac{a^{4}}{a^{2}+2bca^{2}}+\frac{b^{4}}{b^{2}+2acb^{2}}+\frac{c^{4}}{c^{2}+2abc^{2}}\geq \frac{1}{1+2abc(a+b+c)}\geq \frac{1}{1+\frac{2.9(a^{2}+b^{2}+c^{2})^{2}}{27}}\geq \frac{3}{5}$
1,$A=a+b+c-(\frac{2ab^{3}}{a+2b^{3}}+\frac{2bc^{3}}{b+2c^{3}}+\frac{2ca^{3}}{c+2a^{3}})$$a+2b^{3}=a+b^{3}+b^{3}\geq
3b^{2}\sqrt[3]{a}\rightarrow \frac{2ab^{3}}{a+2b^{3}}\leq \frac{2}{3}.\sqrt[3]{ab.ab.b}\leq \frac{2}{9}(2ab+b)$tương tự cuối cùng được: $A\geq a+b+c-\frac{2}{9}(2ab+2bc+2ca+a+b+c)\geq 3-\frac{2}{9}.(2.\frac{9}{3}+3)=1$2,$A=\frac{a^{4}}{a^{2}+2bca^{2}}+\frac{b^{4}}{b^{2}+2acb^{2}}+\frac{c^{4}}{c^{2}+2abc^{2}}\geq \frac{1}{1+2abc(a+b+c)}\geq \frac{1}{1+\frac{2.9(a^{2}+b^{2}+c^{2})^{2}}{27}}\geq \frac{3}{5}$