Prove that $(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$ Let be $a,b,c,x,y,z>0$ such that $ax\ge \sqrt{(b^2+c^2)(y^2+z^2)}$. Prove that
$$(ay-bx)^2+(az-cx)^2\ge (bz-cy)^2$$
I tried to expand
$$a^2(y^2+z^2)+x^2(b^2+c^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
Here my idea was to use the condition after the means inequality:
$$a^2(y^2+z^2)+x^2(b^2+c^2) \ge 2ax\sqrt{(b^2+c^2)(y^2+z^2)}$$
$$\ge 2(b^2+c^2)(y^2+z^2)$$
but it's not good enough to prove the question
$$2(b^2+c^2)(y^2+z^2)+2bcyz\ge b^2z^2+c^2y^2+2abxy+2acxz$$
is not true when $a$ and $x$ can be very big.
Thank you for your help.
 A: Using Cauchy-Schwarz:
$$
\begin{aligned} \left[\left(\frac{c}{a}\right)^2+\left(\frac{b}{a}\right)^2\right]\cdot \left[(ay-bx)^2+(cx-az)^2\right]&\geq \left[\frac{c}{a}(ay-bx)+\frac{b}{a}(cx-az)\right]^2\\
&=(cy-bz)^2\\
\end{aligned}
$$
and similarly
$$
\begin{aligned} \left[\left(\frac{z}{x}\right)^2+\left(\frac{y}{x}\right)^2\right]\cdot \left[(bx-ay)^2+(az-cx)^2\right]&\geq (bz-cy)^2\\
\end{aligned}
$$
Multiplying the two inequalities:
$$\left[\left(\frac{c}{a}\right)^2+\left(\frac{b}{a}\right)^2\right]\cdot \left[\left(\frac{z}{x}\right)^2+\left(\frac{y}{x}\right)^2\right]\cdot \left[(ay-bx)^2+(cx-az)^2\right]^2 \geq (bz-cy)^4$$
and notice that using the condition:
$$\left[\left(\frac{c}{a}\right)^2+\left(\frac{b}{a}\right)^2\right]\cdot \left[\left(\frac{z}{x}\right)^2+\left(\frac{y}{x}\right)^2\right]=\frac{(b^2+c^2)(y^2+z^2)}{a^2x^2}\leq 1$$
It follows that:
$$\left[(ay-bx)^2+(cx-az)^2\right]^2 \geq (bz-cy)^4$$
which is equivalent with the inequality to prove.
A: Let $\frac{b}{a}=p$, $\frac{c}{a}=q$, $\frac{y}{x}=u$ and $\frac{z}{x}=v$.
Thus, the condition it's $$(p^2+q^2)(u^2+v^2)\leq1$$ and we need to prove that:
$$(u-p)^2+(v-q)^2\geq(pv-qu)^2,$$ which is true by C-S twice:
$$(u-p)^2+(v-q)^2=\sqrt{\left((u-p)^2+(v-q)^2\right)^2}\geq$$
$$\geq\sqrt{\left((v-q)^2+(p-u)^2\right)(p^2+q^2)\cdot\left((v-q)^2+(p-u)^2\right)(u^2+v^2)}\geq$$
$$\geq\sqrt{(vp-qp+pq-uq)^2(vu-qu+pv-uv)^2}=(pv-qu)^2.$$
