Ta có: ({{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)+{{log }_{4ab+1}}left( 2a+2b+1 right)=2)
(Leftrightarrow {{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)+frac{1}{{{log }_{2a+2b+1}}left( 4ab+1 right)}=2.)
Có: ({{left( 2a right)}^{2}}+{{b}^{2}}ge 2.2a.bLeftrightarrow 4{{a}^{2}}+{{b}^{2}}ge 4ab.)
(Rightarrow 4{{a}^{2}}+{{b}^{2}}+1ge 4ab+1.)
Dấu “=” xảy ra (Leftrightarrow 2a=b.)
Theo giả thiết ta có:
(left{ begin{array}{l}a > 0\b > 0end{array} right. Rightarrow left{ begin{array}{l}2a + 2b + 1 > 1\4a + 1 > 1end{array} right.) (Rightarrow left{ begin{array}{ccccc}log _{2a + 2b + 1}left( {4{a^2} + {b^2} + 1} right) ge {log _{2a + 2b + 1}}left( {4ab + 1} right)\{log _{2a + 2b + 1}}left( {2a + 2b + 1} right) = frac{1}{{{{log }_{2a + 2b + 1}}left( {4ab + 1} right)}}end{array} right..)
Áp dụng bất đẳng thức Cauchy ta có:
(begin{align} & {{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)+frac{1}{{{log }_{2a+2b+1}}left( 4ab+1 right)}le {{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)+frac{1}{{{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)} \ & le 2.sqrt{{{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right).frac{1}{{{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)}}=2. \end{align})
Dấu “=” xảy ra (Leftrightarrow left{ begin{align} & 2a=b \ & {{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)=frac{1}{{{log }_{2a+2b+1}}left( 4{{a}^{2}}+{{b}^{2}}+1 right)} \end{align} right.)
(begin{array}{l}
Leftrightarrow left{ begin{array}{l}
2a = b\
{log _{2a + 2b + 1}}left( {4{a^2} + {b^2} + 1} right) = 1
end{array} right. Leftrightarrow left{ begin{array}{l}
2a = b\
{log _{3b + 1}}left( {2{b^2} + 1} right) = 1
end{array} right.\
Leftrightarrow left{ begin{array}{l}
b = 2a\
2{b^2} + 1 = 3b + 1
end{array} right. Leftrightarrow left{ begin{array}{l}
b = 2a\
2{b^2} – 3b = 0
end{array} right. Leftrightarrow left{ begin{array}{l}
b = 2a\
left[ begin{array}{l}
b = 0;;left( {ktm} right)\
b = frac{3}{2};;left( {tm} right)
end{array} right.
end{array} right. Leftrightarrow left{ begin{array}{l}
a = frac{3}{4};;left( {tm} right)\
b = frac{3}{2}
end{array} right..
end{array})
Vậy (a+2b=frac{3}{4}+3=frac{15}{4}.)
Chọn A.