Let $a, b,$ and $c$ are positive real numbers such that $a + b + c \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$ Prove that $$\frac{a + b - c}{a^3 + b^3 + abc} + \frac{b + c - a}{b^3 + c^3 + abc} + \frac{c + a - b}{c^3 + a^3 + abc}\leq 1$$

notice we have $a^2bc+ab^2c+abc^2 \geq ab+bc+ca$ $a^3+b^3+abc \geq a^2b+ab^2+abc$ $c(a^3+b^3+abc) \geq a^2bc+ab^2c+abc^2 \geq ab+bc+ca$ $a^3+b^3+abc \geq \frac{ab+bc+ca}{c}$ analogously for other case $LHS \leq \frac{1}{ab+bc+ca}(2ac+2ab+2bc-a^2-b^2-c^2)\leq \frac{(ab+bc+ca)}{ab+bc+ca}=1$ since $ab+bc+ca \leq a^2+b^2+c^2$

Note that if $a+b-c,b+c-a,c+a-b$ are all non-negative, then in light of $x^3+y^3 \geq xy(x+y)$ we obtain $\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{1}{a+b+c} \cdot \dfrac{\displaystyle 2 \sum ab-\sum a^2}{abc} \leq \dfrac{\displaystyle \sum ab}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyle \sum a} \leq 1,$ as desired. Now, if one of $a+b-c,b+c-a,c+a-b$ is negative, WLOG $a+b-c<0$, the other two must be positive as $(a+b-c)+(b+c-a)=2b>0$ and $(a+b-c)+(c+a-b)=2a>0$. Hence, $\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{b+c-a}{b^3+c^3+abc}+\dfrac{c+a-b}{c^3+a^3+abc} \leq \dfrac{b+c-a}{bc(a+b+c)}+\dfrac{c+a-b}{ca(a+b+c)}=$ $=\dfrac{2ab+ac+bc-a^2-b^2}{abc(a+b+c)} \leq \dfrac{ac+bc}{abc(a+b+c)} < \dfrac{ab+bc+ca}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyle \sum a} \leq 1,$ as desired. Equality holds, for example, if $a=b=c=1$.

This was also second problem of the forth Romanian JBMO TST. My solution was the same as @above.

Straightforward problem. Notice that the given condition can be rewritten as $abc(a+b+c) \ge ab+bc+ca$. Moreover, it is easy to see that $a^3+b^3 \ge ab^2+ab^2$, since this is the same as $(a+b)(a-b)^2 \ge 0$, which is obviously true. Hence \begin{align*} \frac{a + b - c}{a^3 + b^3 + abc} + \frac{b + c - a}{b^3 + c^3 + abc} + \frac{c + a - b}{c^3 + a^3 + abc} \le \\ \frac{a+b-c}{ab^2+a^2b+abc} + \frac{b+c-a}{bc^2+cb^2+abc} + \frac{c+a-b}{c^2a+ca^2+abc} = \\ \frac{ac+bc-c^2}{abc(a+b+c)} +\frac{ab+ac-a^2}{abc(a+b+c)} + \frac{bc+ab-b^2}{abc(a+b+c)} \le \\ \frac{2(ab+bc+ca) - (a^2+b^2+c^2)}{ab+bc+ca} = \\ =2 - \frac{a^2+b^2+c^2}{ab+bc+ca} \le 1 \end{align*}where we used the well-known fact that $a^2+b^2+c^2 \ge ab+bc+ca$. The value $1$ can be achieved by taking $a=b=c=1$.

Kimchiks926 wrote: Straightforward problem. Notice that the given condition can be rewritten as $abc(a+b+c) \ge ab+bc+ca$. Moreover, it is easy to see that $a^3+b^3 \ge ab^2+ab^2$, since this is the same as $(a+b)(a-b)^2 \ge 0$, which is obviously true. Hence \begin{align*} \frac{a + b - c}{a^3 + b^3 + abc} + \frac{b + c - a}{b^3 + c^3 + abc} + \frac{c + a - b}{c^3 + a^3 + abc} \le \\ \frac{a+b-c}{ab^2+a^2b+abc} + \frac{b+c-a}{bc^2+cb^2+abc} + \frac{c+a-b}{c^2a+ca^2+abc} = \\ \frac{ac+bc-c^2}{abc(a+b+c)} +\frac{ab+ac-a^2}{abc(a+b+c)} + \frac{bc+ab-b^2}{abc(a+b+c)} \le \\ \frac{2(ab+bc+ca) - (a^2+b^2+c^2)}{ab+bc+ca} = \\ =2 - \frac{a^2+b^2+c^2}{ab+bc+ca} \le 1 \end{align*}where we used the well-known fact that $a^2+b^2+c^2 \ge ab+bc+ca$. The value $1$ can be achieved by taking $a=b=c=1$. if $a+b-c,b+c-a,c+a-b$ are all non-negative, then \begin{align*} \frac{a + b - c}{a^3 + b^3 + abc} + \frac{b + c - a}{b^3 + c^3 + abc} + \frac{c + a - b}{c^3 + a^3 + abc} \le \\ \frac{a+b-c}{ab^2+a^2b+abc} + \frac{b+c-a}{bc^2+cb^2+abc} + \frac{c+a-b}{c^2a+ca^2+abc} \end{align*}

Orestis_Lignos wrote: Note that if $a+b-c,b+c-a,c+a-b$ are all non-negative, then in light of $x^3+y^3 \geq xy(x+y)$ we obtain $\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{1}{a+b+c} \cdot \dfrac{\displaystyle 2 \sum ab-\sum a^2}{abc} \leq \dfrac{\displaystyle \sum ab}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyle \sum a} \leq 1,$ as desired. Now, if one of $a+b-c,b+c-a,c+a-b$ is negative, WLOG $a+b-c<0$, the other two must be positive as $(a+b-c)+(b+c-a)=2b>0$ and $(a+b-c)+(c+a-b)=2a>0$. Hence, $\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{b+c-a}{b^3+c^3+abc}+\dfrac{c+a-b}{c^3+a^3+abc} \leq \dfrac{b+c-a}{bc(a+b+c)}+\dfrac{c+a-b}{ca(a+b+c)}=$ $=\dfrac{2ab+ac+bc-a^2-b^2}{abc(a+b+c)} \leq \dfrac{ac+bc}{abc(a+b+c)} < \dfrac{ab+bc+ca}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyle \sum a} \leq 1,$ as desired. Equality holds, for example, if $a=b=c=1$. Why do you need to consider both cases? does some inequality doesn't hold on one case or the other

I fakesolved this We, assume, that $a=b=c,$ then $3a \ge \frac{3}{a} \implies 3a^2 \ge 3 \implies a^2 \ge 1.$ Now, we choose the minimum value of $a,$ note that this is a fake solve. So, this becomes, $a=b=c=1 \implies \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1.$

The original problem is a special case of the following generalization where $$n=3$$.

Generalization 1 Let $a_{1},a_{2},\cdots,a_{n}$ be positive reals ($n\geq 3$) such that $\sum_{cyc}{a_{1}}\geq \sum_{cyc}{\dfrac{1}{a_{1}}}$. Then prove that $$\sum_{cyc-k}{\left(\dfrac{\dfrac{\prod{a_{1}}}{a_{k+1}a_{k}}+\dfrac{\prod{a_{1}}}{a_{k}a_{k-1}}-\dfrac{\prod{a_{1}}}{a_{k+1}a_{k+2}}}{\dfrac{\prod{a_{1}}}{a_{k}}\left(\sum_{cyc}{a_{1}}\right)}\right)}\leq 1$$ (İndexs are on modulo n.)