Câu hỏi 87
Chứng minh rằng nếu $x, y, z \gt 0$ thì
$$
\frac{(x+1)(y+1)^2}{3\sqrt[3]{z^2x^2}+1} + \frac{(y+1)(z+1)^2}{3\sqrt[3]{x^2y^2}+1} + \frac{(z+1)(x+1)^2}{3\sqrt[3]{y^2z^2}+1} \geq x+y+z+3.
$$
$$
\frac{(x+1)(y+1)^2}{3\sqrt[3]{z^2x^2}+1} + \frac{(y+1)(z+1)^2}{3\sqrt[3]{x^2y^2}+1} + \frac{(z+1)(x+1)^2}{3\sqrt[3]{y^2z^2}+1} \geq x+y+z+3.
$$