Solutions

We can prove the first inequality with a proof by
contradiction.

$$\begin{aligned} x^{2} + xy + y^{2} \leq 0 & \implies x^{2} + 2xy + y^{2} - xy \leq 0 \\ & \implies {(x + y)}^{2} - xy \leq 0 \\ & \implies {(x + y)}^{2} \leq xy \end{aligned}$$

Now, if $x > 0, y < 0$, we then obtain $xy < 0$. Consequently, we would
obtain the following contradiction.

$$\begin{aligned} {(x + y)}^{2} \leq xy < 0 & \implies {(x + y)}^{2} < 0 \end{aligned}$$

Notice, for the second inequality, that the polynomial can be factored.
Based on problem 1.4,

$$\begin{aligned} \frac{x^{5} - y^{5}}{x - y} = x^{4} + x^{3}y + x^{2}y^{2} + xy^{3} + y^{4} \end{aligned}$$

If $\bm{x - y < 0}$:

$$\begin{aligned} x - y < 0 & \implies x < y \\ & \implies x^{5} < y^{5} && (6.1) \\ & \implies x^{5} - y^{5} < 0 \\ & \implies \frac{x^{5} - y^{5}}{x - y} > 0 \end{aligned}$$

On the other hand, if $\bm{x - y > 0}$

$$\begin{aligned} x - y > 0 & \implies x > y \\ & \implies x^{5} > y^{5} && (6.1) \\ & \implies x^{5} - y^{5} > 0 \\ & \implies \frac{x^{5} - y^{5}}{x - y} > 0 \end{aligned}$$

As seen above, in both cases the fraction is greater than 0 and thus so
too is our polynomial.