3. Solve the equation $\frac{1}{2} x^{2}+y^{2}+\frac{3}{2} z^{2}+t^{2}+\frac{1}{3}=x y+y z+z t+t$.
Solution. Multiply everything by 2 and rewrite the equality as

$$
\left(x^{2}-2 x y+y^{2}\right)+\left(y^{2}-2 y z+z^{2}\right)+\left(t^{2}-2 z t+z^{2}\right)+\left(t^{2}-2 t+1\right)+z^{2}=\frac{1}{3}
$$

or

$$
(x-y)^{2}+(y-z)^{2}+(t-z)^{2}+(1-t)^{2}+z^{2}=\frac{1}{3}
$$

We will prove the auxiliary inequality: $a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}(a+b+c)^{2}$, with equality holding only when $a=b=c$. Indeed, this is equivalent to the inequality $3 a^{2}+3 b^{2}+3 c^{2} \geqslant a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$, or $2 a^{2}+2 b^{2}+2 c^{2} \geqslant 2 a b+2 b c+2 c a$. The latter inequality, in turn, is the sum of three Cauchy inequalities: $a^{2}+b^{2} \geqslant 2 a b$, $b^{2}+c^{2} \geqslant 2 b c, c^{2}+a^{2} \geqslant 2 c a$. For equality to hold, it is necessary that $a=b, b=c, c=a$.

Notice that by the proven inequality $(t-z)^{2}+(1-t)^{2}+z^{2} \geqslant \frac{1}{3}$, since $(t-z)+$ $(1-t)+z=1$. Moreover, it is clear that $(x-y)^{2}+(y-z)^{2} \geqslant 0$. Therefore, the original equality can only hold when $x-y=y-z=0$ and $t-z=z=1-t$ (in that order, so that the sum equals 1). Solving this system, we find $x=y=z=1 / 3$, $t=2 / 3$