Let $x,y,z$ be real numbers in the interval $[-1,2]$ such that $x+y+z=0$. Prove that $\sum_{cyc}\sqrt{\frac{(2-x)(2-y)}{(2+x)(2+y)}}\ge 3$ Let $x,y,z$ be real numbers in the interval $[-1,2]$ such that $x+y+z=0$. Prove that
$ \sqrt{\frac{(2-x)(2-y)}{(2+x)(2+y)}}+\sqrt{\frac{(2-y)(2-z)}{(2+y)(2+z)}}+\sqrt{\frac{(2-z)(2-x)}{(2+z)(2+x)}}\ge 3$
My working: let $\frac{2-x}{2+x}=a^2,\frac{2-y}{2+y}=b^2,\frac{2-z}{2+z}=c^2$
$\implies a,b,c\in[0,\sqrt 3], \sum\frac{1-a^2}{1+a^2}=0\text{ or } 3+a^2+b^2+c^2-a^2b^2-b^2c^2-c^2a^2-3a^2b^2c^2=0$ and to prove $ab+bc+ca\ge3$
 A: As you suggested, let $p=a+b+c$, $q=ab+bc+ca$, $r=abc$. We use contradiction: if $q<3$, then $f(p,q,r)=3+a^2+b^2+c^2-a^2b^2-b^2c^2-c^2a^2-3a^2b^2c^2=3+p^2-2q-q^2+2pr-3r^2>0$ and thus leads to a contradiction. If $p,q$ is fixed, we must have $f$ takes its minimum when $r$ is at its extreme. Consider $F(X)=x^3-p^2X+qX$. This is a cubic curve, and the three roots is it intersect with $y=r$. So if $r$ takes the extreme values, we must have: either $a,b,c$ gets to the left, right walls ($0$ and $\sqrt{3}$), or gets to upper, lower walls (two of $a,b,c$ are equal). So we only need to prove the case with one of $a,b,c$ is zero, or $\sqrt{3}$, or $a=b$. That is, for this inequality (and this TYPE of inequalities) we only need to consider the special cases.
Case 1: $a=\sqrt{3}$.
Then $p=(\sqrt{3}+b+c)$, $q=\sqrt{3}(b+c)+bc$, $r=\sqrt{3}bc$. Let $s=b+c$ and $t=bc$, so $f(p,q,r)=6-2(b^2+c^2)-4b^2c^2=6-2s^2+4t-4t^2$. $q<3$ means $\sqrt{3} s+t< 3$ . To minimize $f$, we need $s$ to be maximized, so $s<\sqrt{3}-\frac{1}{\sqrt{3}}t$. So $f=6-2s^2+4t-4t^2>6-2(\sqrt{3}-\frac{1}{\sqrt{3}}t)^2+4t-4t^2=8t-\frac{14}{3}t^2$. Notice that $s^2>4t$ so we have $2\sqrt{t}\le s<\sqrt{3}-\frac{1}{\sqrt{3}}t$ and thus we can conclude that $t<1$. So $8t-\frac{14}{3}t^2\ge 0$ and thus $f>0$.
Case 2: $a=0$.
Then $p=b+c$, $q=bc<3$, $r=0$, so we have $p^2\ge 4q$, and thus $f=3+p^2-2q-q^2\ge 3+4q-2q-q^2=3+2q-q^2=(3-q)(1+q)>0$
Case 3: $a=b$
We have $f=3+2a^2-a^4+(1-2a^2-3a^4)c^2$. The condition $q<3$ transformed to $a(a+2c)<3$. Notice that this $f$ takes the minimum value, the $c^2$ (thus $c$) must take extreme values, so $c=0,c=\sqrt{3}$ or $c<\frac{3}{2a}-\frac{a}{2}$. Take $c=\frac{3}{2a}-\frac{a}{2}$, substitute it in, we have
$$f>3+2a^2-a^4+(1-2a^2-3a^4)(\frac{3}{2a}-\frac{a}{2})^2 = \frac{3(3-a^2)(a-1)^2(a+1)^2(a^2+1)}{4a^2}\ge 0$$
Thus $f>0$.
