Question 125
A simple but very useful inequality is $x^{2} \geq 0$, for every real $x$. To prove it, it suffices to study separately the following possibilities: $x\gt 0, x\lt 0$ or $x=0$. Indeed, a positive real number multiplied by a positive real number is positive, a negative real number multiplied by another negative real number is also positive and, finally, $0 \cdot 0=0$. From this inequality we can derive other, less elementary ones. For example, for any real numbers $x$ and $y$ it is true that $x^{2}+x y+y^{2} \geq 0$, because
$$
\begin{aligned}
x^{2}+x y+y^{2} & =\left(x^{2}+\frac{2 x y}{2}+\frac{y^{2}}{4}\right)+3\left(\frac{y^{2}}{4}\right) \\
& =\left(x+\frac{y}{2}\right)^{2}+3\left(\frac{y}{2}\right)^{2} \\
& \geq 0
\end{aligned}
$$
In the last inequality we used that $\left(x+\frac{y}{2}\right)^{2} \geq 0$ and $\left(\frac{y}{2}\right)^{2} \geq 0$.
We will now see an application of these inequalities. Let $a$ and $b$ be real numbers such that $a^{3}-b^{3}=2$ and $a^{5}-b^{5} \geq 4$.
a. Knowing that for any reals $x$ and $y$ we have $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$, verify that $a\gt b$.
b. Verify that $a^{2}+b^{2} \geq 2$.
$$
\begin{aligned}
x^{2}+x y+y^{2} & =\left(x^{2}+\frac{2 x y}{2}+\frac{y^{2}}{4}\right)+3\left(\frac{y^{2}}{4}\right) \\
& =\left(x+\frac{y}{2}\right)^{2}+3\left(\frac{y}{2}\right)^{2} \\
& \geq 0
\end{aligned}
$$
In the last inequality we used that $\left(x+\frac{y}{2}\right)^{2} \geq 0$ and $\left(\frac{y}{2}\right)^{2} \geq 0$.
We will now see an application of these inequalities. Let $a$ and $b$ be real numbers such that $a^{3}-b^{3}=2$ and $a^{5}-b^{5} \geq 4$.
a. Knowing that for any reals $x$ and $y$ we have $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$, verify that $a\gt b$.
b. Verify that $a^{2}+b^{2} \geq 2$.