1) Tìm min:
[TEX]A=(x+1)^2.(\frac{1}{x^2}+\frac{4}{x}+1)[/TEX] với x>0.
[TEX]B=\frac{a^2}{1-b}+\frac{b^2}{a-1}[/TEX] với a;b >1.
2) Tìm max:
[TEX]C=(x+1)(x-1)(x+3)(3-x)[/TEX]
1 Áp dụng bđt cosi
[TEX]\begin{array}{l}{\left( {x + 1} \right)^2} \ge 4x\\\left( {\frac{1}{{{x^2}}} + 1} \right) + \frac{4}{x} \ge \frac{6}{x}\\ = > VT \ge 24\\ = \Leftrightarrow x = 1\end{array}[/TEX]
2 áp dụng bđt cosi
[TEX]\begin{array}{l}VT = \frac{{{{\left( {a - 1} \right)}^2} + 2\left( {a - 1} \right) + 1}}{{b - 1}} + \frac{{{{\left( {b - 1} \right)}^2} + 2\left( {b - 1} \right) + 1}}{{a - 1}} \ge 4\left( {\frac{{a - 1}}{{b - 1}} + \frac{{b - 1}}{{a - 1}}} \right) \ge 8\\ = \Leftrightarrow a = b = 2\end{array}[/TEX]
3.
[TEX]\begin{array}{l}VT = \left( {{x^2} - 1} \right)\left( {9 - {x^2}} \right) \le \frac{{{{\left( {{x^2} - 1 + 9 - {x^2}} \right)}^2}}}{4} = 16\\ = \Leftrightarrow x = \pm \sqrt 5 \end{array}[/TEX]