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若 ${\displaystyle a,b,c}$ 满足 ${\displaystyle a={\log }_{3}4+{\log }_{5}4,b={\log }_2(4-b),c=2^{4-c}}$，已知 ${\displaystyle 2^8>3^5,4^{10}<5^9}$, 则

• A. ${\displaystyle b+c>2a}$
• B. ${\displaystyle bc<a^2}$
• C. ${\displaystyle 2^c-{\log }_{2}b>2a}$
• D. ${\displaystyle b{\log }_{2}c+c{\log }_{2}b<a^2}$

记 ${\displaystyle f(x)=x-{\log }_2(4-x)}$，则它单增，且 ${\displaystyle f(b)=f(4-c)=0}$,故 ${\displaystyle b+c=4}$.而 ${\displaystyle {\displaystyle f\left(\frac{3}{2}\right)={\log }_2\frac{4\sqrt{2}}{5}>0,f(1)=1-{\log }_{2}3<0}}$，故 ${\displaystyle 1<b<{\displaystyle \frac{3}{2}}}$.
另一方面，根据提示有$$\begin{array}{r}\{\begin{array}{rl}& 4^4>3^5,\\ & 4^{10}<5^9,\end{array}\Rightarrow \{\begin{array}{rl}& {\log }_{3}4>\frac{5}{4},\\ & {\log }_{4}5>\frac{10}{9}.\end{array}\end{array}$$

• A 因为$$\begin{array}{r}a>\frac{5}{4}+{\log }_{5}4>\frac{10}{9}+{\log }_{5}4>{\log }_{4}5+{\log }_{5}4>2,\end{array}$$
  故${\displaystyle b+c=4<2a}$, A错误.
• B 由基本不等式得 ${\displaystyle bc\le 4=4<a^2}$, B正确.
• C ${\displaystyle 2^c-{\log }_{2}b=2^{4-b}-{\log }_{2}b}$ 关于 ${\displaystyle b}$ 单减，故左边大于 ${\displaystyle {\displaystyle 2^{\frac{5}{2}}-{\log }_2\frac{3}{2}=4\sqrt{2}+1-{\log }_{2}3>6.6-{\log }_{2}3}}$.而根据刚才提示的结果，右边小于 ${\displaystyle 2{\log }_{3}4+2\cdot {\displaystyle \frac{10}{9}=4{\log }_{3}2+1.8}}$.这样只需要说明 ${\displaystyle 4.8>4{\log }_{3}2+{\log }_{2}3}$.右侧是一个对勾函数，容易知道 ${\displaystyle 2>{\log }_{2}3>{\displaystyle \frac{3}{2}}}$，因此$$4{\log }_{3}2+{\log }_{2}3\le max\{4,\frac{25}{6}\}<4.8,$$ 从而C正确.
• D 由对数求和不等式 $$\begin{array}{r}b{\log }_{2}c+c{\log }_{2}b=-b{\log }_2\frac{1}{c}-c{\log }_2\frac{1}{b}<-(b+c){\log }_2\frac{1+1}{b+c}=4<a^2,\end{array}$$ 故D正确.

因此答案是BCD.