Why does this Olympiad inequality proving technique (Isolated Fudging) work? A few years ago, I was in a math olympiad training camp and they taught us a technique to prove inequalities. I just came across it again recently. However, I am not able to understand why it works. So, here is how it goes. Suppose, you want to prove
$$ \frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{b+a} \geq \frac{3}{2}.$$
What you do instead is to find an $\alpha$ such that
$$\frac{a}{b+c} \geq \frac{3}{2}\frac{a^\alpha}{a^\alpha+b^\alpha+c^\alpha}. \tag{1}\label{eq1}$$ 
The technique is primarily meant to find such an $\alpha$ (In an actual olympiad, this would be rough work and once you "know" $\alpha$, you would be supposed to prove the new inequality using standard techniques- Cauchy Schwarz, Hölder's...). To find $\alpha$, we set $b=c=1$. Now, we want to prove 
$$\frac{a}{2} \geq \frac{3}{2} \frac{a^\alpha}{a^\alpha +2}$$
$$\Leftrightarrow a^{\alpha+ 1}- 3a^\alpha + 2 a \geq 0$$
Now, we differentiate (wrt a) the equation on the left-hand side and set it equal to zero for a=1. You get
$$\alpha + 1 - 3\alpha + 2 =0$$
$$\Rightarrow \alpha= 3/2$$
My question is why does this procedure work? When does it work? I understand that we are somehow setting the minima of Eq. \eqref{eq1}, but how does it all work out at $a=b=c=1$? I remember (maybe incorrectly) that for the inequality
$$ \sqrt{\frac{a}{b+c}}+ \sqrt{\frac{b}{a+c}}+ \sqrt{\frac{c}{b+a}} \geq 2$$
you need to use $b=1, c=0$. Why and what's the general rule here?
 A: It not always works.
More exactly, we not always can find this trick during a competition without computer.
For example, there is the following estimation (Ji Chen):

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$.
Prove that:
$$\frac{1}{2+a^2+b^2}\leq\frac{3(6a^2+b^2+c^2+2ab+2ac+4bc)}{32(a^2+b^2+c^2+ab+ac+bc)}.$$

This estimation gives $$\sum_{cyc}\frac{1}{2+a^2+b^2}\leq\sum_{cyc}\frac{3(6a^2+b^2+c^2+2ab+2ac+4bc)}{32(a^2+b^2+c^2+ab+ac+bc)}=\frac{3}{4}$$ and we proved that $$\sum_{cyc}\frac{1}{2+a^2+b^2}\leq\frac{3}{4}.$$
It seems as proof in one line, but we need to prove the Ji Chen's inequality and to find it, which is just impossible during the competition.
By the way, the last inequality we can prove by another ways (the best of them it's  uvw, I think).
The inequality $$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2$$ we can prove by your trick:
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2,$$ but I think, much more better to get this estimation by AM-GM:
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2$$ without to look for $\alpha$, for which
$$\sqrt{\frac{a}{b+c}}\geq\frac{2a^{\alpha}}{a^{\alpha}+b^{\alpha}+c^{\alpha}}.$$
Also, we need to check $a=b=1$ and $c=0$ if we need to find some estimation, because the equality in the inequality $$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2$$ occurs in this case.
Also, for any $n\geq2$ by Karamata we obtain:
$$\sqrt[n]{\frac{a}{b+c}}\geq\sqrt{\frac{a^{\frac{2}{n}}}{b^{\frac{2}{n}}+c^{\frac{2}{n}}}}\geq \frac{2a^{\frac{2}{n}}}{a^{\frac{2}{n}}+b^{\frac{2}{n}}+c^{\frac{2}{n}}},$$ which in the general gives:
$$\sum_{cyc}\sqrt[n]{\frac{a}{b+c}}\geq\sum_{cyc}\frac{2a^{\frac{2}{n}}}{a^{\frac{2}{n}}+b^{\frac{2}{n}}+c^{\frac{2}{n}}}=2.$$
