Mathematical Reflections 2017, Issue 3 - Problem J410

Problem:
Let $a,b,c,d$ be real numbers such that $a^2 \leq 2b$ and $c^2<2bd$. Prove that $$x^4+ax^3+bx^2+cx+d>0$$ for all $x \in \mathbb{R}$.

Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania

Solution:
Since $a^2 \leq 2b$ and $c^2<2bd$, we have $b \geq 0$ and $d>0$. Moreover,
$$x^4+ax^3+bx^2+cx+d=\left(x^2+\dfrac{a}{2}x\right)^2+\left(b-\dfrac{a^2}{4}\right)x^2+cx+d.$$ Clearly, $\left(x^2+\dfrac{a}{2}x\right)^2 \geq 0$. Since $b-\dfrac{a^2}{4}>0$ and the discriminant of $\left(b-\dfrac{a^2}{4}\right)x^2+cx+d$ is $$c^2-4\left(b-\dfrac{a^2}{4}\right)d=c^2-4bd+a^2d<-2bd+a^2d=d(a^2-2b) \leq 0,$$ then $$\left(b-\dfrac{a^2}{4}\right)x^2+cx+d>0$$ for all $x \in \mathbb{R}$, which gives $x^4+ax^3+bx^2+cx+d>0$ for all $x \in \mathbb{R}$.