Problem 9.5. (20 points) Real numbers $a, b, c, d$ are such that $a+b=\frac{9}{c-d}$ and $c+d=$ $=\frac{25}{a-b}$. What is the smallest value that the quantity $a^{2}+b^{2}+c^{2}+d^{2}$ can take?
# Answer: 34.

Solution. If the given equalities are multiplied by the denominators of the corresponding fractions and added, we get $2(a c-b d)=34$. We will prove that $a^{2}+b^{2}+c^{2}+d^{2} \geqslant 2(a c-b d)$. This follows from $a^{2}+c^{2} \geqslant 2 a c$ (equivalent to $(a-c)^{2} \geqslant 0$) and $b^{2}+d^{2} \geqslant-2 b d$ (equivalent to $\left.(b+d)^{2} \geqslant 0\right)$. Therefore, $a^{2}+b^{2}+c^{2}+d^{2} \geqslant 34$.

Equality is achieved if all the above inequalities turn into equalities, i.e., when $a=c$ and $b=-d$. Substituting these relations into the equalities given in the problem, it is not difficult to find suitable values, for example, $a=4, b=-1, c=4, d=1$.

Another solution. Let $a+b$ be denoted by $x$, and $a-b$ by $y$. We get $a=\frac{1}{2}(x+y)$ and $b=\frac{1}{2}(x-y)$, which gives

$$
a^{2}+b^{2}=\frac{(x+y)^{2}}{4}+\frac{(x-y)^{2}}{4}=\frac{x^{2}+y^{2}}{2}
$$

On the other hand, from the condition we have $c-d=\frac{9}{x}$ and $c+d=\frac{25}{y}$, from which $c=\frac{1}{2}\left(\frac{9}{x}+\frac{25}{y}\right)$ and $d=\frac{1}{2}\left(\frac{9}{x}-\frac{25}{y}\right)$. Similarly, we get

$$
c^{2}+d^{2}=\frac{1}{4}\left(\frac{9}{x}+\frac{25}{y}\right)^{2}+\frac{1}{4}\left(\frac{9}{x}-\frac{25}{y}\right)^{2}=\frac{1}{2}\left(\frac{9^{2}}{x^{2}}+\frac{25^{2}}{y^{2}}\right)
$$

Adding, we have

$$
a^{2}+b^{2}+c^{2}+d^{2}=\frac{1}{2}\left(x^{2}+\frac{9^{2}}{x^{2}}+y^{2}+\frac{25^{2}}{y^{2}}\right)
$$

Noting that $x^{2}+9^{2} / x^{2} \geqslant 2 \cdot 9$ (which follows from $(x-9 / x)^{2} \geqslant 0$) and similarly $y^{2}+25^{2} / y^{2} \geqslant 2 \cdot 25$, we get

$$
a^{2}+b^{2}+c^{2}+d^{2} \geqslant \frac{1}{2}(2 \cdot 9+2 \cdot 25)=9+25=34
$$

Equality can only be achieved if all inequalities turn into equalities. From $x-9 / x=0$ we get $x= \pm 3$, and from $y-25 / y=0$ we get $y= \pm 5$, from which it is not difficult to obtain suitable examples of the values of the original variables.

Remark. From the given solutions, it can be understood that the expression takes the value 34 only for four sets of $(a, b, c, d)$:

$$
(4,-1,4,1) ; \quad(-1,4,-1,-4) ; \quad(1,-4,1,4) ; \quad(-4,1,-4,-1)
$$

## Criteria

Points for the estimate and example are summed:

+17 p. Estimate - it is proven that $a^{2}+b^{2}+c^{2}+d^{2} \geqslant 34$.

In the absence of a complete proof of the estimate, the highest suitable criterion is used:

+3 p. The work contains the equality $2\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a+b)^{2}+(a-b)^{2}+(c+d)^{2}+(c-d)^{2}$.

+3 p. The work contains the equality $a c-b d=17$ or $2 a c-2 b d=34$.

+0 p. The work contains the equality $a d-b c=8$ or $2 a d-2 b c=16$.

+0 p. The problem is solved under the assumption that the numbers $a, b, c, d$ are integers.

+3 p. Example - suitable numbers $a, b, c, d$ are provided for which $a^{2}+b^{2}+c^{2}+d^{2}=34$ (or it is proven that such exist).

The following advancement is not evaluated:

0 p. Only the answer is provided.