Inequality of logarithm function

[anemone]

Prove that, for all real $a,\,b,\,c$ such that $a+b+c=3$, the following inequality holds:

$\log_3(1+a+b)\log_3(1+b+c)\log_3(1+c+a)\le 1$

 

[MarkFL]

My solution:

Spoiler

Let the objective function be:

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

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

\(\displaystyle a=b=c=1\)

And:

\(\displaystyle f(1,1,1)=1\)

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

\(\displaystyle (a,b,c)=(3,0,0)\)

We find:

\(\displaystyle f(3,0,0)=0\)

Thus, we conclude:

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

 

[anemone]

My solution:

Spoiler

Let the objective function be:

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

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

\(\displaystyle a=b=c=1\)

And:

\(\displaystyle f(1,1,1)=1\)

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

\(\displaystyle (a,b,c)=(3,0,0)\)

We find:

\(\displaystyle f(3,0,0)=0\)

Thus, we conclude:

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

Very good job, MarkFL!

 

[anemone]

My solution, guided by one inequality expert:

Spoiler

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

\(\displaystyle \begin{align*}\sqrt[3]{\log_3(1+a+b)\log_3(1+b+c)\log_3(1+c+a)}&\le \frac{\log_3(1+a+b)+\log_3(1+b+c)+\log_3(1+c+a)}{3}\,\,\tiny{\text{by the AM-GM inequality}}\\&= \log_3(1+a+b)^{\frac{1}{3}}+\log_3(1+b+c)^{\frac{1}{3}}+\log_3(1+c+a)^{\frac{1}{3}}\\&=\log_3((1+a+b)(1+b+c)(1+c+a))^{\frac{1}{3}}\\&\le \log_3\left(\frac{1+a+b+1+b+c+1+c+a}{3}\right)\,\,\small{\text{again by the AM-GM inequality}}\\&=\log_3 \left(\frac{9}{3}\right)\\&=\log_3 3\\&=1\end{align*}\)