MHB Inequality of logarithm function

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that, for all real $a,\,b,\,c$ such that $a+b+c=3$, the following inequality holds:

$\log_3(1+a+b)\log_3(1+b+c)\log_3(1+c+a)\le 1$
 
Mathematics news on Phys.org
My solution:

Let the objective function be:

$$f(a,b,c)=\log_3(1+a+b)\log_3(1+b+c)\log_3(1+a+c)$$

Using cyclic symmetry, we find that the extremum must occur for:

$$a=b=c=1$$

And:

$$f(1,1,1)=1$$

If we pick another point on the constraint, such as:

$$(a,b,c)=(3,0,0)$$

We find:

$$f(3,0,0)=0$$

Thus, we conclude:

$$f(a,b,c)\le1$$
 
MarkFL said:
My solution:

Let the objective function be:

$$f(a,b,c)=\log_3(1+a+b)\log_3(1+b+c)\log_3(1+a+c)$$

Using cyclic symmetry, we find that the extremum must occur for:

$$a=b=c=1$$

And:

$$f(1,1,1)=1$$

If we pick another point on the constraint, such as:

$$(a,b,c)=(3,0,0)$$

We find:

$$f(3,0,0)=0$$

Thus, we conclude:

$$f(a,b,c)\le1$$

Very good job, MarkFL!
 
My solution, guided by one inequality expert:

We're asked to prove $\log_3(1+a+b)\log_3(1+b+c)\log_3(1+c+a)\le 1$, we could, since the RHS of the inequality is a $1$, by taking the cube root to both sides of the inequality and get:

$\sqrt[3]{\log_3(1+a+b)\log_3(1+b+c)\log_3(1+c+a)}\le \sqrt[3]{1}=1$

Therefore

$$\begin{align*}\sqrt[3]{\log_3(1+a+b)\log_3(1+b+c)\log_3(1+c+a)}&\le \frac{\log_3(1+a+b)+\log_3(1+b+c)+\log_3(1+c+a)}{3}\,\,\tiny{\text{by the AM-GM inequality}}\\&= \log_3(1+a+b)^{\frac{1}{3}}+\log_3(1+b+c)^{\frac{1}{3}}+\log_3(1+c+a)^{\frac{1}{3}}\\&=\log_3((1+a+b)(1+b+c)(1+c+a))^{\frac{1}{3}}\\&\le \log_3\left(\frac{1+a+b+1+b+c+1+c+a}{3}\right)\,\,\small{\text{again by the AM-GM inequality}}\\&=\log_3 \left(\frac{9}{3}\right)\\&=\log_3 3\\&=1\end{align*}$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
4
Views
2K
Replies
10
Views
3K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
4
Views
2K
Back
Top