\(log_{a^2}\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}log_a\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}\left[log_aa^3-log_a\sqrt[5]{b^3}\right]=\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)\)

\(\Rightarrow\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)=3\)

\(\Rightarrow log_ab=-5\)

\(P=3log_{a^2b}a-\dfrac{3}{4}log_a2.log_2\left(\dfrac{a}{b}\right)\)

\(=\dfrac{3}{log_a\left(a^2b\right)}-\dfrac{3}{4.log_2a}.\left(log_2a-log_2b\right)\)

\(=\dfrac{3}{log_aa^2+log_ab}-\dfrac{3}{4.log_2a}.log_2a+\dfrac{3}{4}.\dfrac{log_2b}{log_2a}\)

\(=\dfrac{3}{2+3}-\dfrac{3}{4}+\dfrac{3}{4}.log_ab=\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{9}{4}=\dfrac{21}{10}\)

\(B=\left(\log b_a+\log_ba+2\right)\left(\log b_a-\log b_{ab}\right)-1=\left(\log b_a+\frac{1}{\log b_a}+2\right)\left(\log b_a.\log_ba-\left(\log_{ab}b.\log_ba\right)\right)-1\)

   \(=\frac{\log^2_ab+2\log_ab+1}{\log_ab}\left(1-\log_{ab}a\right)-1=\frac{\left(\log_ab+1\right)^2}{\log_ab}\left(1-\frac{1}{\log_aab}\right)-1\)

  \(=\frac{\left(\log_ab+1\right)^2}{\log_ab}\left(1-\frac{1}{1+\log_ab}\right)-1=\frac{\left(\log_ab+1\right)^2}{\log_ab}.\frac{\log_ab}{1+\log_ab}-1=\log_ab+1-1=\log_ab\)

\(P=log_{\dfrac{\sqrt{a}}{b}}a+log_{\dfrac{\sqrt{a}}{b}}\sqrt[3]{b}=log_{\dfrac{\sqrt{a}}{b}}a+\dfrac{1}{3}log_{\dfrac{\sqrt{a}}{b}}b\)

\(=\dfrac{1}{log_a\dfrac{\sqrt{a}}{b}}+\dfrac{1}{3.log_b\dfrac{\sqrt{a}}{b}}=\dfrac{1}{log_a\sqrt{a}-log_ab}+\dfrac{1}{3\left(log_b\sqrt{a}-log_bb\right)}\)

\(=\dfrac{1}{\dfrac{1}{2}-2}+\dfrac{1}{3\left(\dfrac{1}{4}-1\right)}=-\dfrac{10}{9}\)

\(=\left(\log_ab+\log_ba+2\right)\left(1-\log_{ab}a\right)-1\)

\(=\left(\log_ab+\log_ba+2\right)\left(1-\frac{1}{1+\log_ab}\right)-1\)

\(=\frac{1}{1+\log_ab}\left(\log_ab+\log_ba+2\right)-1\)

\(=\frac{1}{1+\log_ab}\left[\left(\log_ab+\log_ba+2\right)-1-\log_ab\right]\)

\(=\frac{1}{1+\log_ab}\left(\log_ab+\log^2_ba\right)=\log_ab\)

a) Áp dụng công thức: \(\log_ab.\log_bc=\log_ac\)

b) Vì \(\dfrac{1}{\log_{a^k}b}=\dfrac{1}{\dfrac{1}{k}\log_ab}=\dfrac{k}{\log_ab}\) nên biểu thức vế trái bằng:

\(VT=\dfrac{1}{\log_ab}\left(1+2+...+n\right)\)

\(=\dfrac{1}{\log_ab}.\dfrac{n\left(n+1\right)}{2}=VP\)

Ta có : 

\(a^{\log_bc}=c^{\log_ba}\Rightarrow a^{\log_bc}+c^{\log_ab}=c^{\log_ba}+c^{\log_ab}\ge2\sqrt{c^{\log_ba}.c^{\log_ab}}=2\sqrt{c^{\log_ba+\log_ab}}\) (1)

Vì \(a,b>1\) nên áp dụng BĐT Cauchy cho 2 số không âm \(\log_ba\) và \(\log_ab\), ta được :

\(\log_ab+\log_ba\ge2\sqrt{\log_ab.\log_ba}=2\)  (2)

Từ (1) và (2) \(\Rightarrow a^{\log_bc}+b^{\log_ab}\ge2\sqrt{c^2}=2c\)

hay \(\Rightarrow a^{\log_bc}+c^{\log_ab}\ge2c\)

Chứng minh tương tự ta được :

                           \(a^{\log_bc}+b^{\log_ca}\ge2a\)

                           \(b^{\log_ca}+c^{\log_ab}\ge2b\)

\(\Rightarrow2\left(a^{\log_bc}+b^{\log_ca}+c^{\log_ab}\right)\ge2\left(a+b+c\right)\)

hay : 

              \(a^{\log_bc}+b^{\log_ca}+c^{\log_ab}\ge a+b+c\)  (*)

Mặt khác theo BĐT Cauchy ta có : \(a+b+c\ge3\sqrt[3]{abc}\)  (2*)

Từ (*) và (2*) ta có : 

                        \(a^{\log_bc}+b^{\log_ca}+c^{\log_ab}\ge3\sqrt[3]{abc}\)