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I was see the Isolated Fudging technical here and found it interesting.

So far, I have done the problem in the post $$\sqrt{\frac{a}{b+c}} \ge \frac{ka^x}{a^x+b^x+c^x}$$ 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. But is it always $\dfrac{a^x}{a^x+b^x+c^2x}$?

I find this $$\frac{(2a+b+c)\sqrt{a^2+bc}}{\sqrt{(b^2+ca)(c^2+ab)}} +\frac{(2b+c+a)\sqrt{b^2+ca}}{\sqrt{(c^2+ab)(a^2+bc)}}+\frac{(2c+a+b)\sqrt{c^2+ab}}{\sqrt{(a^2+bc)(b^2+ca)}}\ge 8$$ true for $a,b,c \ge 0$. Equality case $(0;t;t)$. Obviously, $\dfrac{a^x}{a^x+b^x+c^x}$ doesn't work for $a=0$, then I try $$\frac{(2a+b+c)\sqrt{a^2+bc}}{\sqrt{(b^2+ca)(c^2+ab)}} \ge \frac{k(b^x+c^x)}{a^x+b^x+c^x}$$ When I set $a=0, b=c$ then $k=2$. But if $b=0, a=c$ then $k=3$, so that I don't know how to choose.

Also, another estimate $$\frac{(2a+b+c)\sqrt{a^2+bc}}{\sqrt{(b^2+ca)(c^2+ab)}} \ge \frac{k((ab)^x+(ac)^x)+m(bc)^x}{(ab)^x+(bc)^x+(ca)^x}$$ choose $a=0,b=c$ we get $m=2$. Choose $a=c,b=0$ then we get $k=3$. To find $x$, set $b=0,c=1$ then we can't find $x$. Set $b=c=1$ $$f(a)=2\sqrt{a^2+1}-\frac{6a^x+2}{2a^x+1} \ge 0$$ $$f'(a)=\frac{1}{\sqrt{a^2+1}}+\frac{6xa^{x-1}(2a^x+1)-(6a^x+2)2xa^{x-1}}{(2a+1)^2} $$ Obviously, we have to set $f'(0)=0$ to find $x$, which is impossible.

Can anyone help me with this Isolated Fudging for equality case $(0;t;t)$ and further, two equality cases $(t;t;t)$ and $(0;t;t)$. Thanks a lot!

30 Anh Ti 711
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1 Answers1

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Some thoughts.

We may use this form $$\frac{(2a+b+c)\sqrt{a^2+bc}}{\sqrt{(b^2+ca)(c^2+ab)}} \ge \frac{56a^2 - 8a(b + c) + 32bc}{7(a^2 + b^2 + c^2) + 2(ab + bc + ca)}.$$

River Li
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    Can you tell me the way you find the coefficient please? – 30 Anh Ti 711 Jan 17 '24 at 12:06
  • Yes, I also wanted to know. This form with negative coefficients on the numerator seems new to me. – youthdoo Apr 10 '25 at 02:45
  • @youthdoo Sometimes you can use the conditions of equality. Sometimes it is not easy to find the case that the numerator is positive. – River Li Apr 10 '25 at 05:53
  • Currently do you know any articles about isolated fudging? I am planning to write one based on the techniques and possible motivations that I summarized and inferred. – youthdoo Oct 08 '25 at 02:51
  • @youthdoo I knew this trick from https://yufeizhao.com/olympiad/wc08/ineq.pdf – River Li Oct 08 '25 at 02:57
  • @youthdoo I want to discuss the uvw related issues. Do you think in uvw related theorems, e.g. Tejs' theorem, we can replace uvw notation with pqr notation? In uvw notation, $a+b+c = 3u, ab + bc + ca = 3v^2, abc = w^3$, and for reals, saying that $3v^2$ can be negative, and denoting $f(w^3)$. In pqr notation, $a+b+c=p, ab + bc + ca = q, abc = r$ without these issues e.g. $f(r)$. I don't know why the author of uvw material chooses uvw notation. Any advantage in using uvw notation? – River Li Oct 14 '25 at 05:02
  • Perhaps someone says homogenization. But using pqr, there is no homogenization issue in proving uvw theorems, right? E.g. $pq - 3r$ is homogeneous v.s. $9uv^2 - 3w^3$ is homogeneous. – River Li Oct 14 '25 at 05:02
  • I also prefer $pqr$ notation. Perhaps one other benefit of $uvw$ notation is that when $a=b=c$ also $u=v=w$, so it might be easier to spot whether something is obviously wrong by summing the coefficients. For example, it is obviously wrong that$$u^6+7u^2v^4\ge9w^6.$$But in $pqr$ notation this is less obvious. – youthdoo Oct 14 '25 at 06:28
  • @youthdoo Thanks. It makes sense. – River Li Oct 14 '25 at 07:10
  • @youthdoo In my opinion, sometimes, pqr (using basic relations among $p, q, r$) is better than uvw (with Tejs'stheorem etc.). See [this question] (https://math.stackexchange.com/questions/5089092/prove-sqrt4-fracxyz-sqrt4-fraczyx-sqrt4-fracxzy-ge-3). To check $b=c=1$, we need to prove that $\left(\frac{a+2}{3}\right)^8(a^2+2)\geq(2a^2+1)a^2$. Do you have elegant way to prove it? – River Li Oct 16 '25 at 02:03
  • @youthdoo Yes, you can prove it by giving $\left(\frac{a+2}{3}\right)^8(a^2+2)- (2a^2+1)a^2 = f(a)(a - 1)^2$ where all the coefficients of $f(a)$ are non-negative. But if you want to verify the above by hand, it involves a little bit calculation. – River Li Oct 16 '25 at 02:03
  • What about take logarithm? – youthdoo Oct 16 '25 at 15:01
  • @youthdoo Thanks. A little bit better I think. Some calculations are required (by hand). Any way, the pqr method is better here (it is obvious for AM-GM after using $p^2\ge 3q$). – River Li Oct 17 '25 at 00:17