Cho $\large a>0, b>0$ thỏa mãn $\large \log_{4a+5b+1}(16a^2+b^2+1)+\lo

Cho $\large a>0, b>0$ thỏa mãn $\large \log_{4a+5b+1}(16a^2+b^2+1)+\log_{8ab+1}(4a+5b+1)=2$. Giá trị của $\large a+2b$ bằng 

• A. A. $\large 9$
• B. B. $\large 6$
• C. C. $\large \dfrac{27}{4}$
• D. D. $\large \dfrac{20}{3}$

Chọn C

Theo bất đẳng thức Cauchy với $\large a>0, b>0$, ta có: 

$\large 16a^2+b^2+1\geq \sqrt{16a^2b^2}+1=8ab+1\Rightarrow 16a^2+b^2+1\geq 8ab+1$ (*) 

Do $\large 4a+5b+1>1$ nên từ (*) ta có: 

$\large \log_{4a+5b+1}(16a^2+b^2+1)+\log_{8ab+1}(4a+5b+1)\geq \log_{4a+5b+1}8ab+1+\log_{8ab+1}4a+5b+1$ 

$\large \Rightarrow \log_{4a+5b+1}16a^2+b^2+1+\log+{8ab+1}4a+5b+1\geq \log_{4a+5b+1}8ab+1+\dfrac{1}{\log_{4a+5b+1}8ab+1}$ 

Mặt khác: $\large 4a+5b+1>1, \,\, 8ab+1>1\Rightarrow \log_{4a+5b+1}8ab+1+\dfrac{1}{\log_{4a+5b+1}8ab+1}\geq 2$

Suy ra: $\large \log_{4a+5b+1}16a^2+b^2+1+\log_{8ab+1}4a+5b+1\geq 2$

Đẳng thức xảy ra khi $\large \left\{\begin{align}& 16a^2=b^2\\& 4a+5b+1=8ab+1\\& a,b>0\\\end{align\right. $ $\large \Leftrightarrow \left\{\begin{align}& b=4a\\& 2b^2-6b=0\\& a,b>0\\\end{align}\right.$ $\large \Leftrightarrow \left\{\begin{align}& a=\dfrac{3}{4}\\& b=3\\\end{align}\right.$ 

Vậy $\large a+2b=\dfrac{27}{4}$