<s>[INST] Olympiad inequality: is this reasoning sound? I am trying to show that for $a,b,c>0,\;abc=1:$
$$\underbrace{\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}}_{X}\geq \frac{3}{2}$$
This problem is from the Zhautykov Olympiad of 2008.
Attempt:
If $X\geq \frac{3}{2}$ is satisfied for all $a,b,c$ satisfying the initial conditions then by letting $a\mapsto b$ and $b\mapsto a$, the following inequality must also be satisfied:
$$\underbrace{\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)}}_{Y}\geq\frac{3}{2}$$
The reasoning can be reversed, thus $X\geq\frac{3}{2}\iff Y\geq\frac{3}{2}$. Now, in any case:
$$X+Y=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 3\sqrt[3]{\frac{1}{(abc)^2}}=3$$
By AM-GM. Hence either $X\geq \frac{3}{2}$ or $Y\geq \frac{3}{2}$ is true, and therefore they are both true.
Is this logic valid? It seems too easy.
 [/INST] No, your reasoning is wrong. The inequality 
$$X+Y \geq 3$$ is ok. But you did not proof that if $ a,b,c \gt 0 \land   abc=1 \land X(a,b,c) \ge \frac{3}{2}$ for a special triple $(a,b,c)$ then $Y(a,b,c) \ge \frac{3}{2}$ for the same triple (and vice versa) but you only proved that if $ a,b,c \gt 0 \land   abc=1 \land X(a,b,c) \ge \frac{3}{2}$ for all  triples $(a,b,c)$ then $Y(a,b,c) \ge \frac{3}{2}$ (and vice versa) which is only a change in notation. So you assume $X(a,b,c) \ge \frac{3}{2}$ to prove $X(a,b,c) \ge \frac{3}{2}$
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