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E

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Ta có:

$(2y+3z)(2z+3y) \le \dfrac{25(y+z)^2}{4}$

\Rightarrow $\dfrac{x^2}{(2y+3z)(2z+3y)} \ge \dfrac{4x^2}{25(y+z)^2}$

\Rightarrow $\sum \dfrac{x^2}{(2y+3z)(2z+3y)} \ge \sum \dfrac{4x^2}{25(y+z)^2}=\dfrac{4}{25}(\sum \dfrac{x^2}{(y+z)^2})$

Lại có:

$\dfrac{x^2}{(y+z)^2}+\dfrac{1}{4} \ge \dfrac{x}{y+z}$

\Rightarrow $\sum \dfrac{x^2}{(y+z)^2} \ge \sum \dfrac{x}{y+z} -\dfrac{3}{4} \ge \dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}$

\Rightarrow $\sum \dfrac{x^2}{(2y+3z)(2z+3y)} \ge \dfrac{4}{25}.\dfrac{3}{4}=\dfrac{3}{25}$
 
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