\section*{Problem 1 - 311041}

Determine all real numbers \(k\) for which the following statement (1) is true: (1) For every pair \((a ; b)\) of real numbers \(a, b\), it holds that \(a^{2}+b^{2} \geq k \cdot a b\)
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The statement holds exactly for all \(-2 \leq k \leq 2\). We show this by a complete case distinction based on the size of \(k\):

Case 1: \(k>2\). Then, for \(a=b=1\), the inequality \(a^{2}+b^{2}=2=2 \cdot a b<k \cdot a b\) is satisfied, so statement (1) is false.

Case 2: \(0 \leq k \leq 2\).

Case 2.1: It is \(a b \geq 0\). Then, because \(0 \leq(a-b)^{2}=a^{2}-2 a b+b^{2}\), it follows that \(a^{2}+b^{2} \geq 2 \cdot a b \geq k \cdot a b\).

Case 2.2: It is \(a b<0\). Then \(a^{2}+b^{2} \geq 0 \geq k \cdot a b\). Thus, in both sub-cases, statement (1) is true.

Case 3: \(-2 \leq k<0\).

Case 3.1: It is \(a b \geq 0\). Then \(a^{2}+b^{2} \geq 0 \geq k \cdot a b\).

Case 3.2: It is \(a b<0\). Then, because \(0 \leq(a+b)^{2}=a^{2}+2 a b+b^{2}\), it follows that \(a^{2}+b^{2} \geq(-2) a b \geq k \cdot a b\). Thus, in both sub-cases, statement (1) is true.

Case 4: \(k<-2\). Then, for \(a=1, b=-1\), the inequality \(a^{2}+b^{2}=2=(-2) \cdot a b<k \cdot a b\) is satisfied, so statement (1) is false.

Thus, statement (1) is satisfied exactly for \(-2 \leq k \leq 2\), \(\square\).