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Ta có $\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8} \ge 3\sqrt[3]{\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}.\dfrac{1+a}{8}.\dfrac{1+b}{8}}=\dfrac{3a}{4}$Tương tự$\dfrac{b^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+c}{8}+\dfrac{1+b}{8} \ge \dfrac{3b}{4}$$\dfrac{c^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{1+c}{8}+\dfrac{1+a}{8} \ge \dfrac{3c}{4}$ Cộng theo từng vế ba BĐT trên và rút gọn ta được$\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{b^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{c^3}{\left(1+c\right)\left(1+a\right)} \ge \dfrac{a+b+c}{2}-\dfrac{3}{4} \ge \dfrac{3\sqrt[3]{abc}}{2}-\dfrac{3}{4}= \dfrac{3}{2}-\dfrac{3}{4}= \dfrac{3}{2} $Đẳng thức xảy ra khi$a=b=c=1.$

Ta có $\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8} \ge 3\sqrt[3]{\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}.\dfrac{1+a}{8}.\dfrac{1+b}{8}}=\dfrac{3a}{4}$Tương tự$\dfrac{b^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+c}{8}+\dfrac{1+b}{8} \ge \dfrac{3b}{4}$$\dfrac{c^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{1+c}{8}+\dfrac{1+a}{8} \ge \dfrac{3c}{4}$ Cộng theo từng vế ba BĐT trên và rút gọn ta được$\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{b^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{c^3}{\left(1+c\right)\left(1+a\right)} \ge \dfrac{a+b+c}{2}-\dfrac{3}{4} \ge \dfrac{3\sqrt[3]{abc}}{2}-\dfrac{3}{4}= \dfrac{3}{2}-\dfrac{3}{4}= \dfrac{3}{4} $Đẳng thức xảy ra khi$a=b=c=1.$