<s>[INST] Inequality with parameters Let $2x(ac+bd)+2y(ad-bc)+(a^2+b^2)(1+z)+(c^2+d^2)(1-z) ≥ 0$ for any $a, b, c , d  \in  \mathbb{R}$. Prove that $x^2+y^2+z^2 ≤ 1$.
My attempt:
Take $2$ vectors: ${\vec{u}}=(a;b),{\vec{v}}=(c;d)$.
Then $a c+b d=\left(\vec{u},\vec{v}\right)\nonumber=\,u v\cos\varphi$ is a dot product of these vectors and $a d-b c=[\vec{u},\vec{v}]=u v\sin\varphi$ is a cross product of said vectors.
$\varphi$ is an oriented angle between those vectors measured from $u$ to $v$.
$$a^{2}+b^{2}=u^{2}$$
$$c^{2}+d^{2}=v^{2}$$
As a result, we get: $(1+z)u^{2}+2u v(x\cos\varphi+y\sin\varphi)+(1-z)v^{2}\geq0\forall u,v,\varphi$.
Setting $v=0$ gives $(1+z)\geq0$. Same drill gets us $1-z\geq0$.
Now if $z\ne-1$ the expression is always non-negative iff its discriminant is non-positive i.e.
$$(x\cos\varphi+y\sin\varphi)^{2}+z^{2}\leq1\forall\varphi\in[0;2\pi).$$
Note that $\operatorname*{max}(x\cos\varphi+y\sin\varphi)^{2}=\operatorname*{max}((x;y),(\cos\varphi;\sin\varphi))^{2}=x^{2}+y^{2}$. And thus $x^{2}+y^{2}+z^{2}\leq1$.
The case $z=−1$ gives us:
$$2v(u(x\cos\varphi+y\sin\varphi)+v)\geq0.$$
As we need our equation work for all values of $u,v,\varphi$ we get that
$$\begin{array}{l c r}{{}}&{{}}\\ {{x\cos\varphi+y\sin\varphi\geq-{\frac{v}{u}},u,v\neq0}}.\end{array}$$
As the right side can be made arbitrary close to zero we need to make $x\cos\varphi+y\sin\varphi\geq0\forall\varphi$, which is possible only if $x=y=0$. So in all cases $x^2+y^2+z^2≤1$.
Is my proof correct? Is there any other way to solve this problem?
 [/INST] Letting $c = 1, d = 0$, the hypothesis yields
$$2ax - 2by + (a^2 + b^2)(1+z) + 1 - z \ge 0, \quad \forall a, b\in \mathbb{R}. \tag{1}$$
From (1), clearly we have $z + 1 \ge 0$.
If $z + 1 = 0$, (1) yields
$$2ax - 2by + 2 \ge 0, \quad \forall a, b \in \mathbb{R}$$
which results in $x = y = 0$.
Thus, $x^2 + y^2 + z^2 \le 1$.
If $z + 1 > 0$, (1) is written as: $\forall a, b \in \mathbb{R}$,
$$(z + 1)\left(a + \frac{x}{z+1}\right)^2
+ (z + 1)\left(b - \frac{y}{z+1}\right)^2 + \frac{1 - x^2 - y^2 -z^2}{z + 1}\ge 0. \tag{2}$$
From (2), we have $1 - x^2 - y^2 - z^2 \ge 0$.
We are done.
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