Finding the minimum value of $\log _d a + \log _bd + \log _ac + \log _c b$

Question

The question says to find the minimum value of $$\log _da + \log _bd + \log _ac + \log _cb$$ Given, $a,b,c,d \; \in R^+ -\Bigl(1\Bigl)$
My approach:
I used the AM-GM inequality and so we can see that $$\frac{\log _da + \log _bd + \log _ac + \log _cb}{4} \geq \Biggl(\log _da \cdot \log _bd \cdot \log _ac \cdot \log_cb\Biggl)^\frac{1}{4}$$ So the minimum value is, $$4\Biggl(\log _da \cdot \log _bd \cdot \log _ac \cdot \log _cb\Biggl)^\frac{1}{4}$$ Now, I wrote it as, $$\Biggl(\frac{\log_da}{\log_db} \cdot \frac{\log_ca}{\log_cb}\Biggl)^\frac{1}{4}$$ I am not sure what do to after this point. I also feel like we can use HM inequality but that will only give us, $$\log _da + \log _bd + \log _ac + \log _cb = \log _ad + \log _db + \log _ca + \log _bc $$

Answer 1

Hint.

Assuming positive values, You can use also

$$ \log_y x = \frac{\log x}{\log y} $$

then

$$ \frac{\log a}{\log d}+\frac{\log d}{\log b}+\frac{\log c}{\log a}+\frac{\log b}{\log c}\ge 4 $$

NOTE

Another approach.

Assuming $a\ne 0,b\ne 0,c\ne 0,d\ne 0\;\;$ and calling

$$ f(a,b,c,d) = \frac ad+\frac db+\frac ca+\frac bc $$

we have

$$ \nabla f = \left\{ \begin{array}{c} \frac{1}{d}-\frac{c}{a^2}=0 \\ \frac{1}{c}-\frac{d}{b^2}=0 \\ \frac{1}{a}-\frac{b}{c^2}=0 \\ \frac{1}{b}-\frac{a}{d^2}=0 \\ \end{array} \right. $$

and solving

$$ a = b = c = d \;\;\; \mbox{which gives}\;\;\;\; f(a,b,c,d) \ge 4 $$

because

$$ H = \left( \begin{array}{cccc} \frac{2 c}{a^3} & 0 & -\frac{1}{a^2} & -\frac{1}{d^2} \\ 0 & \frac{2 d}{b^3} & -\frac{1}{c^2} & -\frac{1}{b^2} \\ -\frac{1}{a^2} & -\frac{1}{c^2} & \frac{2 b}{c^3} & 0 \\ -\frac{1}{d^2} & -\frac{1}{b^2} & 0 & \frac{2 a}{d^3} \\ \end{array} \right) $$

and for $a=b=c=d$

$$ H = \left( \begin{array}{cccc} 2 & 0 & -1 & -1 \\ 0 & 2 & -1 & -1 \\ -1 & -1 & 2 & 0 \\ -1 & -1 & 0 & 2 \\ \end{array} \right) $$

with eigenvalues $4,2,2,0$

Answer 2

$$\log _{ d } a+\log _{ b } d+\log _{ a } c+\log _{ c } b\geq 4(\log _{ d } a\cdot \log _{ b } d\cdot \log _{ a } c\cdot \log _{ c } b)^{ \frac { 1 }{ 4 } }=\\ =4(\frac { \ln { a } }{ \ln { d } } \cdot \frac { \ln { d } }{ \ln { b } } \cdot \frac { \ln { c } }{ \ln { a } } \cdot \frac { \ln { b } }{ \ln { c } } )^{ \frac { 1 }{ 4 } }=4$$

Answer 3

Assuming all logarithms are $>0$

$$\dfrac{\log_da+\log_bd+\log_ac+\log_cb}4\ge\sqrt[4]{\log_da\log_bd\log_ac\log_cb}=1$$

using $$\log_xy=\dfrac{\log y}{\log x}$$

the equality occurs if $$\log_da=\log_bd=\log_ac=\log_cb$$

Answer 4

I presume $a,\ldots,d>1$. Then $$\log_a d=\frac{\ln d}{\ln a}$$ etc. Then $$\log_da \cdot \log_bd \cdot \log_ac \cdot \log_cb =\frac{\ln a\ln d\ln c\ln b}{\ln d\ln b\ln a\ln c}=1$$ etc.

Answer 5

By Rearrangement inequality, since $\log$ is an increasing function, we have

$$\log _da + \log _bd + \log _ac + \log _cb=$$ $$=\frac{\log a}{\log d}+\frac{\log d}{\log b}+\frac{\log c}{\log a}+\frac{\log b}{\log c}\ge \frac{\log a}{\log a}+\frac{\log b}{\log b}+\frac{\log c}{\log c}+\frac{\log d}{\log d}=4$$