Chứng minh abc(1 + a^2)(1 + b^2)(1 + c^2) < = 8

abc(1 + a²)(1 + b²)(1 + c²) ≤ 8

⇔ a(1 + a²).b(1 + b²).c(1 + c²) ≤ 8

Ta có: 2a(1 + a²) ≤ \(\frac{{{{\left[ {2a + \left( {1 + {a^2}} \right)} \right]}^2}}}{4} = \frac{{{{\left( {a + 1} \right)}^4}}}{4}\) ⇔ a(1 + a²) ≤ \(\frac{{{{\left( {a + 1} \right)}^4}}}{8}\)

2b(1 + b²) ≤ \(\frac{{{{\left[ {2b + \left( {1 + {b^2}} \right)} \right]}^2}}}{4} = \frac{{{{\left( {b + 1} \right)}^4}}}{4}\)⇔ b(1 + b²) ≤ \(\frac{{{{\left( {b + 1} \right)}^4}}}{8}\)

2c(1 + c²) ≤ \(\frac{{{{\left[ {2c + \left( {1 + {c^2}} \right)} \right]}^2}}}{4} = \frac{{{{\left( {c + 1} \right)}^4}}}{4}\)⇔ c(1 + c²) ≤ \(\frac{{{{\left( {c + 1} \right)}^4}}}{8}\)

Suy ra: a(1 + a²).b(1 + b²).c(1 + c²) ≤ \(\frac{{{{\left( {a + 1} \right)}^4}}}{8}.\frac{{{{\left( {b + 1} \right)}^4}}}{8}.\frac{{{{\left( {c + 1} \right)}^4}}}{8}\)

Mà (a + 1)(b + 1)(c + 1) ≤ \(\frac{{{{\left( {a + 1 + b + 1 + c + 1} \right)}^3}}}{{27}} = \frac{{{{\left( {a + b + c + 3} \right)}^3}}}{{27}} = 8\)

Suy ra: a(1 + a²).b(1 + b²).c(1 + c²) ≤ \(\frac{{{8^4}}}{{8.8.8}} = 8\)

Dấu “=” xảy ra khi: \(\left\{ \begin{array}{l}{\left( {a - 1} \right)^2} = 0\\a = b = c\\a + b + c = 3\end{array} \right. \Leftrightarrow a = b = c = 1\).