Given that $a, b, c$ are positive distinct real numbers such that quadratic expressions $ax^2 + bx + c$, $bx^2 + cx + a$ and $cx^2 + ax + b$ are always non-negative. Then the expression $\frac{a^2 + b^2 + c^2}{ab + bc + ca}$ can never lie in:
Step-by-Step Solution
Key Concept: The non-negativity conditions on all three cyclic quadratics impose constraints via discriminant inequalities that bound the ratio of sum of squares to sum of products.
For a quadratic $px^2 + qx + r$ to be always non-negative, we need $q^2 - 4pr ≤ 0$. Applying this to all three quadratics: $b^2 ≤ 4ac$, $c^2 ≤ 4ab$, and $a^2 ≤ 4bc$. Multiplying all three inequalities gives $(abc)^2 ≤ 64(abc)^2$, which is always satisfied. Adding the three inequalities: $a^2 + b^2 + c^2 ≤ 4(ab + bc + ca)$. Therefore $\frac{a^2 + b^2 + c^2}{ab + bc + ca} ≤ 4$. By Cauchy-Schwarz and the constraint that $a, b, c$ are distinct, we can show this ratio approaches 2 from above but never equals 1. Thus the expression lies in $(1, 4]$ but not in $(-∞, 1]$ or $[4, ∞)$.
Correct Answer: 2,4