Đặt $a=2^x,b=2^y,c=2^z\Rightarrow a,b,c>0 và a^3+b^3+c^3=3$BĐT cần c/m tương đương với
$\frac{a^2}{3-a^2}+\frac{b^2}{3-b^2}+\frac{c^2}{3-c^2}\geq \frac{3}{2}\Leftrightarrow \frac{a^3}{a(3-a^2)}+\frac{b^3}{b(3-b^2)}+\frac{c^3}{c(3-c^2)}\geq \frac{3}{2}$
Theo bđt Cô-si ta có
$2\left[ {a(3-a^2)} \right]^2=2a^2(3-a^2)(3-a^2)\leq (\frac{2a^2+3-a^2+3-a^2}{3})^3=8\Rightarrow a(3-a^2)\leq 2\Rightarrow \frac{a^3}{a(3-a^2)}\geq \frac{a^3}{2}$
Tương tự ta có
$\frac{b^3}{b(3-b^2)}\geq \frac{b^3}{2};\frac{c^3}{c(3-c^2)}\geq \frac{c^3}{2}$
$\Rightarrow\frac{a^3}{a(3-a^2)}+\frac{b^3}{b(3-b^2)}+\frac{c^3}{c(3-c^2)}\geq \frac{1}{2} (a^3+b^3+c^3)=\frac{1}{2}\times 3=\frac{3}{2}$
Dấu "=" xảy ra khi $a=b=c=1$ hay$x=y=z=0$