4. Find the largest $\mathrm{C}$ such that for all $\mathrm{y} \geq 4 \mathrm{x}>0$ the inequality $x^{2}+y^{2} \geq$ C $x y$ holds.
Answer: $17 / 4$.

Sketch of the solution. Let $y=4, x=1.1+16 \geq 4$. So, $\mathrm{C} \leq 17 / 4$.

We will prove that for $\mathrm{C}=17 / 4$ and all $y \geq 4 x>0$, the inequality $x^{2}+y^{2} \geq С x y$ holds. Divide the inequality by $x^{2}>0$ and introduce a new variable $z=y / x \geq 4$. Then, $z^{2}+1 \geq C z \Leftrightarrow(z-C / 2)^{2}+1-C^{2} / 4 \geq 0$. For $z \geq C / 2$, the function $f(z)=z^{2}-C z+1=(z-C / 2)^{2}+1-C^{2} / 4$ is increasing, and thus, from $z \geq 4 \geq C / 2$ it follows that $f(z) \geq f(4)=0$, which is equivalent to the inequality to be proved.