Inspired by Gurido Cuong

We make the substitution $x=ab, y=bc, z=ac$ .

Verify the following relationships:

$xy=ab^{2}c, yz=abc^{2}, xz=a^{2}bc, \sqrt{xyz}=abc, x+y+z=1$

Therefore,

$a+b=\frac{x(y+z)}{\sqrt{xyz}}$

$b+c=\frac{y(x+z)}{\sqrt{xyz}}$

$c+a=\frac{z(x+y)}{\sqrt{xyz}}$

Substitute this into our inequality:

$\frac{3ab+1}{a+b}+\frac{3bc+1}{b+c}+\frac{3ac+1}{a+c}$

$=\sqrt{xyz}(\frac{3x+1}{x(y+z)}+\frac{3y+1}{y(x+z)}+\frac{3z+1}{z(x+y)})$

$=\sqrt{xyz}(\frac{3x+1}{x(1-x)}+\frac{3y+1}{y(1-y)}+\frac{3z+1}{z(1-z)})$

Let's attempt to evaluate the minimum of the function $f(x)=\frac{3x+1}{x(1-x)}$ .

The first derivative of the function is evaluated as $f'(x)=\frac{(3x-1)(x+1)}{x^{2}{(1-x)}^{2}}$ . Setting $f'(x)=0$ , we find that $x=\frac{1}{3}$ or $x=-1$ . We reject the latter since $x$ must be positive.

Consider the shape of the function when graphed. $f(x)\rightarrow \infty$ as $x \rightarrow 0^{+}$ or $x \rightarrow 1^{-}$ . Therefore, we conclude that $f(x)$ does not have a maximum, so the turning point we evaluate must be a minimum instead. This saves us the hassle of evaluating the second derivative.

Therefore, we conclude that the minimum of the function is $9$ , attained when $x=\frac{1}{3}$ . Repeating the same thing to the function in terms of $y$ and $z$ gives us the same results. We find that $x=y=z=\frac{1}{3}$ if equality is to hold. Note also that $x+y+z=1$ , so the constraint of the problem is fulfilled.

Therefore, we have

$\sqrt{xyz}(\frac{3x+1}{x(1-x)}+\frac{3y+1}{y(1-y)}+\frac{3z+1}{z(1-z)})$

$\geq \sqrt{xyz}(9+9+9)$

$= 27abc$ , hence the answer.

Equality holds when $x=y=z=\frac{1}{3}$ , implying $a=b=c=\frac{1}{\sqrt{3}}$

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This problem is motivated by the process of solving This will give you a run for your money by Gurido Cuong, which is an excellent problem. I play around with the conditions and come up with the substitution above.

∵( $\frac{3ab+1}{a+b}$ + $\frac{3bc+1}{b+c}$ + $\frac{3ac+1}{a+c}$ )[(a+b)(3ab+1)+(b+c)(3bc+1)+(a+c)(3ac+1)]≥（3ab+3bc+3ac+3）²=36 ( Cauchy inequality)

（we know ab+bc+ac=1 but 3ab+1,3ac+1,3bc+1 are numerators which makes it difficult for them to interact.In order to break the "barriers" Cauchy inequality is a good choice. I still don't know whether this will work but it's not a bad idea to have a try.)

∴ $\frac{3ab+1}{a+b}$ + $\frac{3bc+1}{b+c}$ + $\frac{3ac+1}{a+c}$ ≥ $\frac{36}{(a+b)(3ab+1)+(b+c)(3bc+1)+(a+c)(3ac+1)}$ …….(*)

(it seems like a monster isn't it? Maybe you'll choose to give up this kind of solution and it's exactly what I thought at first. it's ok .There do be the chances that it's nonsense and it's a tough decision .So it's the atmosphere that I use to decide this kind of decision.In this question,what I've done at first makes "a b c" get together which provides the chances for them to interact.Moreover, the symmetry still exists during the "operation".All above provide a sign of success.So it is worth keeping going.

How to deal with it is the next problem. The fewer structures the beter.Because it makes the problem simple to operate generally. Inspired by this ,I tried to get rid of a+b b+c a+c and replaced them with ab bc ac . )

∵ab+bc+ca=1 ∴a+b= $\frac{1-ab}{c}$ ∴(a+b)(3ab+1)= $\frac{(1-ab)(3ab+1)}{c}$

(after this ,it's obvious that we can apply Gn-An to this to attain its maximum. the direction of the inequality is still right. though we still have no idea what will happen it's not bad to have a try)

∵(1-ab)(3ab+1)= $\frac{1}{3}$ （3-3ab）(1+3ab)≤ $\frac{1}{3}$ ( $\frac{3-3ab+1+3ab}{2}$ )²= $\frac{4}{3}$

Repeating the same thing to the b+c a+c, we can conclude that (*)≥ 36 $\frac{3}{4}$ ( $\frac{1}{c}$ + $\frac{1}{a}$ + $\frac{1}{b}$ ) ^(-1)=27abc

(after doing the same thing it's such a coincidence that ( $\frac{1}{c}$ + $\frac{1}{a}$ + $\frac{1}{b}$ )^(-1) could be abc .How would You know it that If You didn't try )

Equality holds when a=b=c= (1/3)½

hence the largest K is 27

(as stated above it is good to try.If you don't try it's difficult to find the magic behind it .But not All thoughts are worth trying If Time limited.At that Time the atmosphere really helps. )