15. Find the real numbers $a, b, c$ that satisfy the following inequalities: $4(ab + bc + ca) - 1 \geqslant a^2 + b^2 + c^2 \geqslant 3(a^3 + b^3 + c^3) \cdot$ (2005 Australian National Training Team Problem)
15. Since $3\left(a^{3}+b^{3}+c^{3}\right) \geqslant\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)$, from $a^{2}+b^{2}+$ $c^{2} \geqslant 3\left(a^{3}+b^{3}+c^{3}\right)$ we get $a+b+c \leqslant 1$, thus $(a+b+c)^{2} \leqslant 1$, which means
$$a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \leqslant 1$$

Also,
$$4(a b+b c+c a) \geqslant a^{2}+b^{2}+c^{2}+1 \geqslant a^{2}+b^{2}+c^{2}+(a+b+c)^{2}$$

Therefore $\square$
$$2(a b+b c+c a) \geqslant 2\left(-a^{2}+b^{2}+c^{2}\right)$$

That is,
$$a b+b c+c a \geqslant a^{2}+b^{2}+c^{2}$$

But $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, so, $a=b=c$.
Let $a=b=c=x$, then $12 x^{2}-\mathrm{F} \geqslant 3 x^{2} \geqslant 9 x^{3}$, thus $x=\frac{1}{3}$, i.e., $a=b=c=\frac{1}{3}$.