# 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?

• Usually some inequalities have minima/maxima when all the terms are equal or they are extreme. May 4, 2020 at 9:26
• The technique you use is called isolated fudging. It is very useful (but it doesn't always work). See also yufeizhao.com/olympiad/wc08/ineq.pdf and math.stackexchange.com/questions/1370084/…. May 5, 2020 at 8:11
• @Batominovski Well, some years ago, I learned this trick from your 1st link. May 7, 2020 at 4:08
• @Noel Why you deleted the Karamata's tag? You looked for a general rule and I found this rule by Karamata. Also, I really wrote a proof. May 15, 2020 at 6:29
• @MichaelRozenberg I did not understand that completely
– Noel
May 19, 2020 at 5:48

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: $$\frac{1}{2+a^2+b^2}\leq\frac{3(6c^2+a^2+b^2+2ac+2bc+4ab)}{32(a^2+b^2+c^2+ab+ac+bc)}.$$

This estimation gives $$\sum_{cyc}\frac{1}{2+a^2+b^2}\leq\sum_{cyc}\frac{3(6c^2+a^2+b^2+2ac+2bc+4ab)}{32(a^2+b^2+c^2+ab+ac+bc)}=\frac{3}{4}$$ 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: $$\sum_{cyc}\sqrt[n]{\frac{a}{b+c}}\geq\sum_{cyc}\frac{2a^{\frac{2}{n}}}{a^{\frac{2}{n}}+b^{\frac{2}{n}}+c^{\frac{2}{n}}}=2.$$

• This does not answer my question though. I would like to understand how and when the technique works, whereas your answer is about applications of different inequalities
– Noel
May 4, 2020 at 11:09
• @Noel See please better my post. I says that this method does not help in the general and I says about your type inequalities. May 4, 2020 at 12:56
• @Noel See please better my post. I said that this method does not help in the general and I said about your type inequalities. If you want, I can get another examples, when this technique works. My answer on the question:"Why does this technique work?" Because the inequality is very easy. For hard inequalities it's not so useful. But there are examples, when it works. May 15, 2020 at 7:22
• I was just waiting for more answers. I'll accept your answer by the end of the week if there are no more answers.
– Noel
May 19, 2020 at 5:46
I have already done your problem. $$\sqrt{\frac{a}{b+c}} \ge \frac{ka^x}{a^x+b^x+c^x}$$ The equality case is $$(0;t;t)$$. If $$a=0$$, it's obvious. If $$b=0$$, set $$a=c$$ then we have $$\fbox{k=2}$$
To find $$x$$, set $$b=0,c=1$$ $$f(a)=\sqrt{a}-\frac{2a^x}{a^x+1}\ge 0$$ $$f'(a)=\frac{1}{2\sqrt{a}} -\frac{2xa^{x-1}(a^x+1)-2a^x\cdot xa^{x-1}}{(a^x+1)^2}$$ Let $$f'(1)=0$$ then $$\frac{1}{2}-\frac{4x-2x}{2}=0 \iff \fbox{x=1}$$ hence we get the key $$\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$$ which can easily prove by AM GM.