Solution-2

Since the square of a real number is nonnegative, we have

\[\begin{aligned} & \left(x^{2}-1\right)^{2} \geq 0 \\ & x^{4}-2 x^{2}+1 \geq 0 \\ & x^{4}+1 \geq 2 x^{2} \end{aligned}\]

Similarly, we obtain \(y^{4}+1 \geq 2 y^{2}\). Multiplying the last two inequalities together (this is allowed because both sides of both inequalities are nonnegative), we get

\[\left(x^{4}+1\right)\left(y^{4}+1\right) \geq 4 x^{2} y^{2}\]

From the preceding arguments it follows that equality holds if and only if \(\left(x^{2}-1\right)^{2}=0\) and \(\left(y^{2}-1\right)^{2}=0\), that is, \(x^{2}=y^{2}=1\). Thus the equation has four solutions: \((-1;-1)\), \((-1;1)\), \((1;-1)\) and \((1;1)\).

Atrisinājums-2

Tā kā reāla skaitļa kvadrāts ir nenegatīvs, tad

\[\begin{aligned} & \left(x^{2}-1\right)^{2} \geq 0 \\ & x^{4}-2 x^{2}+1 \geq 0 \\ & x^{4}+1 \geq 2 x^{2} \end{aligned}\]

Analoģiski iegūstam, ka \(y^{4}+1 \geq 2 y^{2}\). Sareizinot kopā pēdējās divas nevienādības (to drīkst darīt, jo abu nevienādību abas puses ir nenegatīvas), iegūstam, ka

\[\left(x^{4}+1\right)\left(y^{4}+1\right) \geq 4 x^{2} y^{2}\]

No iepriekš veiktajiem spriedumiem izriet, ka vienādība tiks sasniegta tad un tikai tad, ja \(\left(x^{2}-1\right)^{2}=0\) un \(\left(y^{2}-1\right)^{2}=0\), tas ir, \(x^{2}=y^{2}=1\). Tātad vienādojumam ir četri atrisinājumi: \((-1;-1)\), \((-1;1)\), \((1;-1)\) un \((1;1)\).