6. Determine all real numbers $r$ such that the inequality $a^{3}+a b+b^{3} \geqq a^{2}+b^{2}$ holds for all pairs of real numbers $a, b$ which are greater than or equal to $r$.
SOLUTION. Since

$$
a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right),
$$

we can rewrite the given inequality in the equivalent form

$$
(a+b-1)\left(a^{2}-a b+b^{2}\right) \geqq 0 .
$$

Given that for any real numbers $a, b$,

$$
a^{2}-a b+b^{2}=\left(a-\frac{1}{2} b\right)^{2}+\frac{3}{4} b^{2} \geqq 0
$$

(with equality if and only if $a=b=0$), inequality (1) is satisfied if and only if

$$
a+b \geqq 1 \quad \text { or } \quad a=b=0
$$

From the condition $a+b \geqq 1$, it follows that inequality (1) is guaranteed to be satisfied for all pairs of real numbers $a, b$ that are greater than or equal to $\frac{1}{2}$, so any $r \geqq \frac{1}{2}$ has the required property. However, no $r<\frac{1}{2}$ has this property, because for such an $r$ one can choose positive numbers $a=b=\max \left(r, \frac{1}{3}\right)$, which are indeed greater than or equal to $r$, but do not satisfy inequality (1), as neither $a+b \geqq 1$ nor $a=b=0$ holds for them.

Conclusion. The given problem is satisfied precisely by all real numbers $r \geqq \frac{1}{2}$.

GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Prove that for every odd number $n$ and for any real numbers $a, b$,

$$
a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+\ldots-a b^{n-2}+b^{n-1}\right) .
$$

N2. Plot in the Cartesian coordinate system all points whose coordinates $x, y$ satisfy the relations:
a) $(x-y \geqq 1) \wedge(2 x+y \leqq 1)$,
b) $(\min (x, y) \geqq r) \wedge(\max (x, y) \leqq R)$, where $r<R$ are given real numbers.

N3. Prove that for any positive real numbers $a, b, c$,

$$
\frac{1}{a^{2}-a b+b^{2}}+\frac{1}{b^{2}-b c+c^{2}}+\frac{1}{c^{2}-c a+a^{2}} \leqq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} .
$$

Determine when equality holds. [64-B-I-6]