Problem in a question concerning Geometric progression and quadratic equations

Question

Consider the following problem:-

If $a,b,c,d,p$ are distinct real numbers such that

$(a^2+b^2+c^2)p^2 -2p(ab+bc+cd)+(b^2+c^2+d^2)≤0$ then prove that $a,b,c,d$ are in geometric progression.

The proof is as follows(d part):-

This is all well, but I thought of the problem differently as a quadratic in 'p'. Clearly the coefficient of $x^2$ is $>0$ and thus, for any $D( D<0, D=0 \textrm{ or } D>0)$ ($D$ is discrimant) the quadratic cannot be $≤0$ for all values of $p$. Hence, there is not proof as the given inequality is false.

Hence, the discrepancy between the methods to this problem. Where have I gone wrong?

Answer 1

The question did not require $f(p) \le 0$ for all values of p.

When y = f(p) opens upward and D > 0, the curve will cross the p-axis at two points. Those p that lie between these two points will cause $f(p) \le 0$

Answer 2

In your reasoning should be $\Delta\geq0$, but by C-S we obtain $\Delta\leq0,$

which gives $\Delta=0$ and we obtain a geometric progression again because $$(ab+bc+cd)^2=(a^2+b^2+c^2)(b^2+c^2+d^2)$$ says that $$(a,b,c)||(b,c,d).$$

By the way, $$(ap-b)^2+(bp-c)^2+(cp-d)^2\leq0$$ indeed gives $$ap-b=bp-c=cp-a=0.$$ But the rest is not full.

If $p=0$ thus, $b=c=d=0,$ which is impossible.

Thus, $p\neq 0$ and from here $a$, $b$, $c$ and $d$ are non-zero real numbers.

Thus, indeed, $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$ and we got a geometric progression.