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Áp dụng bất đẳng thức Côsi:
\(P=\left[\frac{a}{\left(a+1\right)^2}+\frac{a+1}{8}+\frac{a+1}{8}+\frac{1}{4a}\right]+\frac{3}{4}\left(a+\frac{1}{a}\right)-\frac{1}{4}\)
\(\ge4\sqrt[4]{\frac{a}{\left(a+1\right)^2}.\frac{a+1}{8}.\frac{a+1}{8}.\frac{1}{4a}}+\frac{3}{4}.2\sqrt{a.\frac{1}{a}}-\frac{1}{4}\)
\(=1+\frac{3}{2}-\frac{1}{4}=\frac{9}{4}\)
Dấu bằng xảy ra khi \(\hept{\begin{cases}a=\frac{1}{a}\\\frac{a}{\left(a+1\right)^2}=\frac{a+1}{8}=\frac{1}{4a}\end{cases}\Leftrightarrow}a=1\)