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!
