Cho ba số thực dương x,y, z thỏa mãn x + y + z bé hơn bằng 1. Chứng minh rằng

Ta có

\[\begin{array}{l}\,\,\,\,\,\,\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right) \ge 512{x^2}{y^2}{z^2}\\ \Leftrightarrow \left( {1 - x} \right)\left( {1 + x} \right)\left( {1 - y} \right)\left( {1 + y} \right)\left( {1 - z} \right)\left( {1 + z} \right) \ge 512{x^2}{y^2}{z^2}\end{array}\]

Do \(x + y + z \le 1\) nên ta có

\(\begin{array}{l}\,\,\,\,\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)\left( {1 + x} \right)\left( {1 + y} \right)\left( {1 + z} \right)\\ \ge \left( {y + z} \right)\left( {z + x} \right)\left( {x + y} \right)\left( {2x + y + z} \right)\left( {x + 2y + z} \right)\left( {x + y + 2z} \right)\,\,\,\,\,\,\,\left( 1 \right)\end{array}\)

Chứng minh được: \(\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) \ge 8xyz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right).\)

Và:

                    \(\begin{array}{l}\,\,\,\left( {2x + y + z} \right)\left( {x + 2y + z} \right)\left( {x + y + 2z} \right)\\ \ge 2\sqrt {x + y} \sqrt {x + z} \,\,\,2\sqrt {y + x} \sqrt {y + z} \,\,\,2\sqrt {z + x} \sqrt {z + y} \\ = 8\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)\\ \ge 8.8xyz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right).\end{array}\)

Từ (1), (2) và (3) suy ra điều phải chứng minh.

Dấu bằng xảy ra khi và chỉ khi \(x = y = z = \frac{1}{3}.\)