<s>[INST] Prove that $a^2+u^2+d^2-b^2-c^2-v^2>-4w^2$ Let $ABCD$ be cyclic quadrilateral of the circle $O$ with:  $$R=w\text{ is radius };AB=a;BC=b;CD=c;DA=d;AC=u;BD=v$$. Prove that $$a^2+u^2+d^2-b^2-c^2-v^2>-4w^2$$


We have $$u=\sqrt{\frac{\left(ac+bd\right)\left(ad+bc\right)}{ab+cd}};v=\sqrt{\frac{\left(ac+bd\right)\left(ab+cd\right)}{ad+bc}}$$ and $$R=\frac{1}{4}\sqrt{\frac{\left(ab+cd\right)\left(ac+bd\right)\left(ad+bc\right)}{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}} \text{for 
 } s=\frac{a+b+c+d}{2}$$
Then by BW and computer we're done. But it's very ugly.I have no idea to solve it without computer. Help me.
 [/INST] Let us denote by $x,y$ the two angles in $A$ delimited by the diagonal $AC$ and the sides $AD$, respectively $AB$. Let us also denote by $s,t$  the two angles in $C$ delimited by the same diagonal $CA$ and the sides $CD$, respectively $CB$. We have
$$
x+y+s+t=\pi\ .
$$
Then we can express all the data in the inequality in terms of $R$ and sine functions in (sums of some of) the variables $x,y,s,t$, for instance, $a/2=R\sin t$, $b/2=R\sin y$, $v/2=R\sin(x+s)=R\sin(y+t)$.
Then $a^2=4R^2\sin^2t=2R^2(1-\cos(2t))$, and similarily for the other squares, so it is useful to introduce $X,Y,S,T$ equal to respectively $2x,2y,2s,2t$ to lower the degrees of the appearing trigonometric functions,
$$
X+Y+S+T=2\pi\ .$$ 
Then we have to show equivalently, step by step:
$$
\begin{aligned}
4R^2 + a^2 + u^2 + d^2 &> c^2 + v^2 + b^2\ ,
\\
1 + \sin^2t + \sin^2(x+s) + \sin^2s &>
    \sin^2x + \sin^2(x+y) + \sin^2y\ ,
\\
2
-\cos T-\cos(X+S)-\cos(S) &>
-\cos X-\cos(X+Y)-\cos(Y) \ ,
\\
2
+\cos X+\cos(X+Y)+\cos(Y) 
 &>
\cos T+\cos(X+S)+\cos(S)\ ,
\\
2
+\cos X+\cos(X+Y)+\cos(Y) 
 &>
\cos (X+Y+S)+\cos(X+S)+\cos S\ ,
\\
2
+\cos X+\cos(X+Y)+\cos(Y) 
 &>
\cos S\Big[\ \cos(X+Y)+\cos X+1\ \Big]
\\
&\qquad-\sin S\Big[\ \sin(X+Y)+\sin X\ \Big] =:E(S,X,Y)\ .
\\[3mm]
&\qquad\text { Here we break the chain of equivalences.}
\\
&\qquad\text { We take the maximum w.r.t. $S$ on the R.H.S. above.}
\\
&\qquad\text { Let us show first:}
\\
(2
+\cos X+\cos(X+Y)+\cos(Y))^2
 &\ge
%(\cos^2 S+\sin^2 S)
%\Big[\ 
\Big(\cos(X+Y)+\cos X+1\Big)^2+\Big(\sin(X+Y)+\sin X\Big)^2
%\ \Big]
\\
&\qquad\text{ i.e. equivalently}
\\
(2
+\underbrace{\cos X+\cos(X+Y)+\cos(Y)}_{=:u})^2
 &\ge
1+1+1+
\underbrace{2\cos X+2\cos(X+Y)+2\cos Y}_{=2u}\ .
\end{aligned}
$$
Above $u\in[-3/2,\ 3]$ (so the above quantity $2-u$ is indeed $>0$, and we could apply the square function in that inequality, obtaining an equivalent inequality,) is a substitute for the sum of cosine functions in $X,X+Y,Y$. The inequality  $(2+u)^2\ge3+2u$ becomes $(1+u)^2\ge0$. This is clear. The strict inequality fails in the case $u=-1$, equivalently either $X=\pi$, or $Y=\pi$. By initial symmetry, we consider only $Y=\pi$.  In this special case, we have to show $2+\cos X>\cos S$. The equality is possible only in case of $X=\pi$. This is a degenerated case with $A=B=D$...
</s>