Chứng minh abc(1 + a^2)(1 + b^2)(1 + c^2) < = 8
Câu hỏi:
Cho các số thực dương a, b, c thỏa mãn a + b + c = 3. Chứng minh
abc(1 + a2)(1 + b2)(1 + c2) ≤ 8.
Trả lời:
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: \(\left\{ \begin{array}{l}{\left( {a - 1} \right)^2} = 0\\a = b = c\\a + b + c = 3\end{array} \right. \Leftrightarrow a = b = c = 1\).
