1,$$\frac{1}{x^3\left(yz+zt+ty\right)}+\dfrac{1}{y^3\left(xz+zt+tx\right)}+\dfrac{1}{z^3\left(xy+yt+tx\right)}+\dfrac{1}{t^3\left(xy+yz+xz\right)}$$
$$
=\dfrac{\dfrac{1}{x^2}}{xyz+xzt+xyt}+\dfrac{\dfrac{1}{y^2}}{xyz+yzt+txy}+\dfrac{\dfrac{1}{z^2}}{xyz+yzt+ztx}+\dfrac{\dfrac{1}{t^2}}{xyt+yzt+txz}$$
$$=\dfrac{\dfrac{1}{x^2}}{\dfrac{xyz}{xyzt}+\dfrac{xzt}{xyzt}+\dfrac{xyt}{xyzt}}+\dfrac{\dfrac{1}{y^2}}{\dfrac{xyz}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{txy}{xyzt}}+\dfrac{\dfrac{1}{z^2}}{\dfrac{xyz}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{ztx}{xyzt}}+\dfrac{\dfrac{1}{t^2}}{\dfrac{xyt}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{txz}{xyzt}}$$
$$=\dfrac{\dfrac{1}{x^2}}{\dfrac{1}{t}+\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\dfrac{1}{y^2}}{\dfrac{1}{t}+\dfrac{1}{x}+\dfrac{1}{z}}+\dfrac{\dfrac{1}{z^2}}{\dfrac{1}{t}+\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\dfrac{1}{t^2}}{\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}}$$
$$\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)^2}{3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)}
$$
$$=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right) \ge \frac{1}{3}.4\sqrt[4]\frac{1}{xyzt}=\frac{4}{3}$$
2,Sử dụng BĐT AM-GM, ta có:
$$\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8} \ge \frac{3a}{4}$$
$$\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{1+c}{8}+\dfrac{1+a}{8}\ge\dfrac{3b}{4}$$
$$\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8}\ge\dfrac{3c}{4}$$
$$\Rightarrow LHS+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}$$
$$LHS \ge \frac{a+b+c}{2}-\frac{6}{8} \ge \frac{3\sqrt[3]{abc}}{2}-\frac{3}{4}=\frac{3}{4}$$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
