\[\frac{6\sqrt{y^{2} + 3}}{y^{2} + 12} \leq 1\]
\[y^{2} + 12 \neq 0\]
\[y - любое\ число.\]
\[\sqrt{y^{2} + 3} > 0\ при\ любом\ y;\]
\[y^{2} + 12 > 0\ при\ любом\ y;\]
\[6\sqrt{y^{2} + 3} \leq y^{2} + 12\]
\[36 \cdot \left( y^{2} + 3 \right) \leq \left( y^{2} + 12 \right)^{2}\]
\[36y^{2} + 108 \leq y^{4} + 24y^{2} + 144\]
\[y^{4} - 12y^{2} + 36 \geq 0\]
\[\left( y^{2} - 6 \right)^{2} \geq 0 - при\ любом\ \text{y.}\]
\[Что\ и\ требовалось\ доказать.\]