$\begin{array}{l}
\frac{a}{{4{b^2}}} + \frac{{2b}}{{{{\left( {a + b} \right)}^2}}} \ge \frac{9}{{4\left( {a + 2b} \right)}}\\
\leftrightarrow \frac{{{a^2} + 2ab}}{{{b^2}}} + \frac{{8ab + 16{b^2}}}{{{{\left( {a + b} \right)}^2}}} \ge 9\\
dat\;\left\{ \begin{array}{l}
x = a + b > 0\\
y = b > 0
\end{array} \right. \to a = x - y > 0\\
\to \frac{{{a^2} + 2ab}}{{{b^2}}} + \frac{{8ab + 16{b^2}}}{{{{\left( {a + b} \right)}^2}}} = \frac{{{{\left( {x - y} \right)}^2} + 2\left( {x - y} \right)y}}{{{y^2}}} + \frac{{8\left( {x - y} \right)y + 16{y^2}}}{{{x^2}}}\\
= \frac{{{x^2} - {y^2}}}{{{y^2}}} + \frac{{8xy + 8{y^2}}}{{{x^2}}} = \frac{{{x^2}}}{{{y^2}}} + \frac{{8y}}{x} + \frac{{8{y^2}}}{{{x^2}}} - 1\\
\cos i:\\
\frac{{{x^2}}}{{{y^2}}} + \frac{8y}{x} + \frac{{8{y^2}}}{{{x^2}}} = \frac{{{x^2}}}{{2{y^2}}} + \frac{{{x^2}}}{{2{y^2}}} + \frac{{4y}}{x} + \frac{{4y}}{x} + \frac{{8{y^2}}}{{{x^2}}} \ge 5\sqrt[5]{{\frac{{{x^4}*{4^2}*8{y^4}}}{{4{y^4}{x^4}}}}} = 10\\
\to dpcm
\end{array}$
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