ta có \(\left(log^b_a+log^a_b+2\right)\left(log^b_a-log_{ab}^b\right).log_b^a-1=\left(log^b_a+log^a_b+2\right)\left(log^b_a.log_b^a-log_{ab}^b.log_b^a\right)-1=\left(log^b_a+log^a_b+2\right)\left(1-\frac{1}{log_b^{ba}}log_b^a\right)-1=\left(log^b_a+log^a_b+2\right)\left(1-\frac{1}{1+log^a_b}log^a_b\right)-1=\left(log^b_a+log^a_b+2\right)\frac{1}{1+log^a_b}-1=\left(log^a_b+\frac{1}{log^a_b}+2\right)\frac{1}{1+log^a_b}-1=\frac{\left(1+log^a_b\right)^2}{log^a_b}\frac{1}{1+log^a}-1=\frac{1+log^a_b}{log_b^a}-1=\frac{1}{log_b^a}\)

 ta có:

\(\left(log^b_a+\frac{1}{log^b_a}+2\right)\left(log^b_a-\frac{1}{log^{ab}_a}\right)log^a_b-1\)\(=\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(log^b_a-\frac{1}{1+log^b_a}\right)log^a_b-1\)\(=\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(1-\frac{log^a_b}{1+log^b_a}\right)-1\)\(==\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(\frac{1}{1+log^b_a}\right)-1=\frac{1+log^b_a}{log^b_a}-1=\frac{1}{log^b_a}\)

Ta thấy rằng do a < b nên \(\log_ab>1\)

Khi đó nếu xét cùng cơ số là b thì : \(\log_a\left(\log_ab\right)>\log_b\left(\log_ab\right)>0\)

Ta cũng có \(\log_ca< 1\) do a < c, suy ra \(0>\log_c\left(\log_ca\right)>\log_b\left(\log_ca\right)\)

Từ đó suy ra :

\(\log_a\left(\log_ab\right)+\log_b\left(\log_bc\right)+\log_c\left(\log_ca\right)>\log_b\left(\log_ab.\log_bc.\log_ca\right)=0\)

Cho \(\log_ab=3;\log_ac=-2\)

1. Với \(x=a^3b^2\sqrt{c}\Rightarrow\log_ax=\log_a\left(a^3b^2\sqrt{c}\right)=\log_aa^3+\log_ab^2+\log_ac^{\frac{1}{2}}\)

             \(=3+2.3+\frac{1}{2}\left(-2\right)=8\)

2. Với \(x=\frac{a^4\sqrt[3]{b}}{c^3}\) \(\Rightarrow\log_a\frac{a^4\sqrt[3]{b}}{c^2}=\log_aa^4+\log_ab^{\frac{1}{3}}+\log_ac^3\)

                                              \(=4+\frac{1}{3}\log_ab+3\log_ac=4+\frac{1}{3}.3+3\left(-2\right)=-1\)

3. Với \(x=\log_a\frac{a^2\sqrt[3]{b}c}{\sqrt[3]{a\sqrt{c}}b^3}\Rightarrow\log_a\frac{a^2b^{\frac{1}{3}}c}{a^{\frac{1}{3}}b^3c^{\frac{1}{6}}}=\log_a\frac{a^{\frac{5}{3}}c^{\frac{5}{6}}}{b^{\frac{8}{3}}}=\log_aa^{\frac{5}{3}}-\log_ab^{\frac{8}{3}}+\log_ac^{\frac{3}{2}}\)

                                                           \(=\frac{5}{3}-\frac{8}{3}\log_ab+\frac{5}{6}\log_ac=\frac{5}{3}-\frac{8}{3}3+\frac{5}{6}\left(-2\right)=-8\)

 ta có:

\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)

\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)

\(=3log^2_2a+3log^a_2+1\)

\(P=\frac{1}{2}log_{\frac{a}{b}}a-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2log_a\frac{a}{b}}-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2\left(1-log_ab\right)}-4log_a\left(a+\frac{b}{4}\right)\)

Ta có: \(a+\frac{b}{4}\ge2\sqrt{\frac{ab}{4}}=\sqrt{ab}\)

\(\Rightarrow log_a\left(a+\frac{b}{4}\right)\le log_a\sqrt{ab}\) (do \(0< a< 1\))

\(\Rightarrow P\ge\frac{1}{2\left(1-log_ab\right)}-4log_a\sqrt{ab}=\frac{1}{2\left(1-log_ab\right)}-2\left(1+log_ab\right)\)

Đặt \(log_ab=x\Rightarrow0< x< 1\) \(\Rightarrow P\ge\frac{1}{2\left(1-x\right)}-2\left(1+x\right)\)

Xét hàm \(f\left(x\right)=\frac{1}{2\left(1-x\right)}-2\left(1+x\right)\) với \(0< x< 1\)

\(f'\left(x\right)=\frac{1}{2\left(1-x\right)^2}-2=0\Leftrightarrow\frac{1-4\left(1-x\right)^2}{2\left(1-x\right)^2}=0\Rightarrow x=\frac{1}{2}\)

Từ BBT ta thấy \(f\left(x\right)_{min}=f\left(\frac{1}{2}\right)=-2\)

\(\Rightarrow P\ge-2\Rightarrow P_{min}=-2\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}log_ab=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\a=\frac{b}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{16}\\b=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow S=\frac{5}{16}\)

Ta có \(\log_ab\ge\log_{a+c}\left(b+c\right)\) với \(1< a\le b\) và \(c\ge0\)

Áp dụng với b = a+1 và c = 1 ta được :

 \(\log_a\left(a+1\right)>\log_{a+1}\left(a+2\right)\)

=> Điều phải chứng minh

Ta có: 

\(\left(b-\dfrac{1}{2}\right)^2\ge0\) <=> \(b^2-b+\dfrac{1}{4}\ge0\) <=>\(b-\dfrac{1}{4}\le b^2\)

Mà : 

a<1 => \(log_a\left(b-\dfrac{1}{4}\right)\ge log_ab^2=2log_ab\)

P=\(log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}log_{\dfrac{a}{b}}b=log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\ge2log_ab-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\)

Đặt t=log_ab

Do b<a<1 => t=log_ab >1

Khi đó \(P\ge2t+\dfrac{t}{2t-2}=f\left(t\right)\). Khảo sát f(t) trên (1;+\(\infty\)) ta đc

P\(\ge\)f(t) \(\ge\) f\(\left(\dfrac{3}{2}\right)\) = \(\dfrac{9}{2}\)