MHB Inequality of logarithm function

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

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

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