Áp dụng bđt sau $3(x^3+y^3+z^3)^2 \ge (x^2+y^2+z^2)$Ta có $3(\frac 1{a^3}+ \frac 1{b^3} + \frac 1{c^3})^2 \ge (\frac{1}{a^2}+ \frac 1{b^2} + \frac 1{c^2})^3 \ge ( \frac 1{ab} + \frac 1{bc} + \frac 1{ca})^3=(\frac{a+b+c}{abc})^3=(\frac{4}{3})^3$$\Rightarrow VT \ge \sqrt{\frac{(\frac 43)^3}{3}}= \frac 38$
Áp dụng bđt sau $3(x^3+y^3+z^3)^2 \ge (x^2+y^2+z^2)
^3$Ta có $3(\frac 1{a^3}+ \frac 1{b^3} + \frac 1{c^3})^2 \ge (\frac{1}{a^2}+ \frac 1{b^2} + \frac 1{c^2})^3 \ge ( \frac 1{ab} + \frac 1{bc} + \frac 1{ca})^3=(\frac{a+b+c}{abc})^3=(\frac{4}{3})^3$$\Rightarrow VT \ge \sqrt{\frac{(\frac 43)^3}{3}}= \frac 38$