Prove that if $x$ and $y$ are both not $0$ Prove that if $x$ and $y$ are both  not $0,$ then 
$$x^4+x^3y+x^2y^2+xy^3+y^4>0$$
I know this seems fairly easy but I'm fairly new to calculus and need some help proving that this is true. Appreciate the help!
 A: Let's divide throughout by $y^4$ (since it is always nonnegative), and then set $z=x/y$ to get the equivalent inequality $$f(z)=z^4+z^3+z^2+z+1>0.$$
To show this, simply note that $f(z)=(z^5-1)/(z-1)$ for $z\neq 1$, and break into two cases when $z\geq1$ and $z<1$, then discuss separately. I'll leave that to you.
A: $$
\begin{align}
x^4+x^3y+x^2y^2+xy^3+y^4
&=\frac12\overbrace{\left(x^4+y^4\right)}^{\substack{\text{$\gt0$ if}\\\text{$(x,y)\ne(0,0)$}}}+\frac12\overbrace{\left(x^2+y^2\right)(x+y)^2}^{\ge0}
\end{align}
$$
A: Another way:
Note that $0\leq x^2(x+y)^2=x^4+2x^3y+x^2y^2$, and similarly $y^4+2xy^3+x^2y^2\geq 0$. Adding these and dividing by $2$, we get $\frac12x^4+x^3y+x^2y^2+xy^3+\frac12y^4\geq 0$. Then add $\frac{x^4+y^4}{2}$, which is positive since $x$ and $y$ are not both $0$.
A: \begin{align}
& x^4 + x^3y + x^2y^2 + xy^3 + y^4 \\[8pt]
= {} & \left( \left( \tfrac x y \right)^4 + \left( \tfrac x y \right)^3 + \left( \tfrac x y \right)^2 + \left( \tfrac x y \right) + 1 \right) y^4 \\[8pt]
= {} & \Big(u^4 + u^3 + u^2 + u + 1\Big) y^4.
\end{align}
Since $y^4>0$ except when $y=0,$ it is enough to look at this $4$th-degree polynomial in $u.$
Recall from algebra that
$$
(u-1)(u^4+u^3+u^2+u+1) = u^5-1.
$$
If you solve $u^5-1=0$ for $u$ you get the $5$th roots of $1$ as solutions. De Moivre's formula says these are
$$
\cos \tfrac {2\pi k} 5 + i\sin \tfrac{2\pi k} 5 \text{ for } k = 0,1,2,3,4.
$$
When $k=0$ this number is $1$ and that is a root of $u-1=0,$ so the other four are the roots of $u^4+u^3+u^2+1=0.$ Since the coefficients of this polynomial are real, the roots must come in complex-conjugate pairs. And here we note that the cases $k=1$ and $k=4$ are complex conjugates of each other, as are the cases $k=2$ and $k=3,$ so we have these roots:
$$
\cos\tfrac{2\pi} 5 \pm i \sin\tfrac {2\pi}5 \text{ and } \cos\tfrac{4\pi}5 \pm i \sin\tfrac{4\pi}5.
$$
Therefore
\begin{align}
& u^4+u^3+u^2+u+1 \\[8pt]
= {} & \left( u - \left( \cos\tfrac{2\pi} 5 + i \sin\tfrac {2\pi}5 \right) \right) \left( u - \left(\cos\tfrac{2\pi} 5 - i \sin\tfrac {2\pi}5\right) \right) \\
& {} \times \left( u - \left( \cos\tfrac{4\pi} 5 + i \sin\tfrac {2\pi}5 \right) \right) \left( u - \left(\cos\tfrac{2\pi} 5 - i \sin\tfrac {4\pi}5\right) \right) \\[8pt]
= {} & \left( u^2 - u\cos\tfrac{2\pi} 5 + 1 \right) \left( u^2 - u \cos\tfrac{4\pi} 5 + 1 \right)
\end{align}
Each of these is a quadratic polynomial whose roots are not real. Each has a graph that is a parabola that opens upward, not downward. Therefore each factor is always positive.
I would call this algebra rather than calculus. I would also guess that most students who have taken the prerequisite courses for calculus would not handle this well at all.
A: Here is a Calculus proof. The result is clear if $x=y\neq 0$. Assume now that $x-y\neq 0$. Let $f(x)=x^5$. Then $f'(x)=5x^4$. By the mean value theorem, one has $$f(x)-f(y)=f'(\xi)(x-y) {\rm ~for~some~}\xi~{\rm between~}x~{\rm and~}y,$$ where $f'(\xi)=5\xi^4>0,$ i.e. $\xi\neq 0$, otherwise $x$ and $y$ must be of different signs, so $f(x)-f(y)\neq 0$ but $f'(0)(x-y)=0$, a contradiction. To prove the original assertion, note that $$x^4+x^3y+x^2y^2+xy^3+y^4=\frac {f(x)-f(y)}{x-y}=f'(\xi)>0,$$ as required. QED
A: Assuming $x=a$ and $y=-b$ with $a\ge b\gt0,$ the left side equals
$$(a^3-b^3)(a-b) + a^2b^2.$$
