Yes, it's just AM-GM.

Here's my solution. \[ \frac{a + bc}{a + a^2} + \frac{b + ca}{b + b^2} + \frac{c + ab}{c + c^2} \]\[ = \sum_{cyc} \frac{1}{a + 1} + \sum_{cyc} \frac{ac}{b(b+1)} \]\[ = \sum_{cyc} \frac{\frac{1}{a+1} + \frac{1}{b+1}}{2} + \sum_{cyc} \frac{\frac{ac}{b(b+1)}+ \frac{bc}{a(a+1)}}{2} \]\[ \ge \sum_{cyc} \frac{1}{\sqrt{(a+1)(b+1)}} + \sum_{cyc} \frac{c}{\sqrt{(a+1)(b+1)}} \]\[ = \sum_{cyc} \frac{c+1}{\sqrt{(a+1)(b+1)}} \]\[ \ge 3 \] CMIIW

$$\sum_{cyc} \frac{a + bc}{a + a^2} = \frac{1}{2}\sum_{cyc} \left(\frac{1}{1+a} + \frac{1}{1+b}+ \frac{bc}{a(1+a)}+\frac{ca}{b(1+b)}\right) \ge\sum_{cyc} \frac{1+c}{\sqrt{(1+a)(1+b)}} \ge 3 $$Very nice.

$$\frac{a+bc}{a+a^2}+\frac{b+ca}{b+b^2}+\frac{c+ab}{c+c^2} \geq \frac{a+ab}{a+a^2}+\frac{b+bc}{b+b^2}+\frac{c+ca}{c+c^2}=\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c} \geq 3$$

Trivial by Jensen's Reasoning: Minimum is clearly established at $a=b=c$ due to convexity of $f''$. Given this, the minimum is $3$. Therefore, the inequality must hold.

mathchampion1 wrote: Trivial by Jensen's Reasoning: Minimum is clearly established at $a=b=c$ due to convexity of $f''$. Given this, the minimum is $3$. Therefore, the inequality must hold. what is $f$?

Well, the idea is that $a=b=c$ has to be an optimum value, because the function is cyclic. This must also be the minimum, shown by Jensen.

mathchampion1 wrote: Well, the idea is that $a=b=c$ has to be an optimum value, because the function is cyclic. This must also be the minimum, shown by Jensen. why not a maximum? what did Jensen say by the way?

For the sake of argument, assume it is optimizable: Function is cyclic, so $a=b=c$ produces an optimum of $3$. This is the only possible optimum, as $a=b=c$ is an if and only if condition. Now, it suffices that $a=1$, $b=2$, and $c=3$ gives a value greater than $3$, so $3$ must be a minimum. Therefore, our bounds must be $\geq 3$. You can use Jensen in this to show that optima actually exist.

mathchampion1 wrote: For the sake of argument: Function is cyclic, so $a=b=c$ produces an optimum of $3$. This is the only possible optimum, as $a=b=c$ is an if and only if condition. Now, it suffices that $a=1$, $b=2$, and $c=3$ gives a value greater than $3$, so $3$ must be a minimum. Therefore, our bounds must be $\geq 3$. you are mixing cyclicity with Jensen...which theorem is this? please just state the theorem.

The theorem that $a=b=c$ optimizes the function?

mathchampion1 wrote: The theorem that $a=b=c$ optimizes the function? the theorem that you are referring to in your pseudo-solution!

what theorem are you talking of? $a=b=c$ optimization is trivial due to the cyclic nature of functions everything else stated in the solution is trivial

mathchampion1 wrote: what theorem are you talking of? $a=b=c$ optimization is trivial due to the cyclic nature of functions everything else stated in the solution is trivial could you please write a mathematical proof stating correctly the theorems and references? Check post 8, that is mathematics. you wrote 10 times more and I am still not convinced!

A little manipulation and then application of the Rearrangements Inequality solves the problem.

Devastator wrote:

what kind of bash did you mean?

Devastator wrote: If $a$,$b$,$c$ are positive reals, prove that $$\frac{a+bc}{a+a^2}+\frac{b+ca}{b+b^2}+\frac{c+ab}{c+c^2} \geq 3$$ The following inequality is also true. ]If $a$,$b$,$c$ are positive reals, prove that \[\frac{a+bc}{a+a^2}+\frac{b+ca}{b+b^2}+\frac{c+ab}{c+c^2} \leq \frac{1}{3}\left(\frac{a^3+b^3+c^3}{abc}\right)^2\]

If $a$,$b$,$c$ are positive reals, prove that \[\frac{1}{3}\left(\frac{a^3+b^3+c^3}{abc}\right)^2\geq \frac{a+bc}{a+a^2}+\frac{b+ca}{b+b^2}+\frac{c+ab}{c+c^2} \geq 3. \]