4. Which is greater: $\sqrt{2016}+\sqrt{2015+\sqrt{2016}}$ or $\sqrt{2015}+\sqrt{2016+\sqrt{2015}}$?
Answer: $\sqrt{2016}+\sqrt{2015+\sqrt{2016}}>\sqrt{2015}+\sqrt{2016+\sqrt{2015}}$.

Solution. First method. Let $a=\sqrt{2016}+\sqrt{2015+\sqrt{2016}}>0, b=\sqrt{2015}+\sqrt{2016+\sqrt{2015}}>$ 0. Then $a^{2}-b^{2}=(\sqrt{2016}+\sqrt{2015+\sqrt{2016}})^{2}-(\sqrt{2015}+\sqrt{2016+\sqrt{2015}})^{2}=\sqrt{2016}-\sqrt{2015}+$ $2 \sqrt{2016 \cdot 2015+2016 \sqrt{2016}}-2 \sqrt{2016 \cdot 2015+2015 \sqrt{2015}}>0$, since $\sqrt{2016}>\sqrt{2015}$ and $2016 \sqrt{2016}>$ $2015 \sqrt{2015}$ (the function $\mathrm{y}=\sqrt{x}$ is increasing). Therefore, $a^{2}>b^{2}$, that is, $a>b$.

Second method. Compare $\sqrt{2016}-\sqrt{2015}$ and $\sqrt{2016+\sqrt{2015}}-\sqrt{2015+\sqrt{2016}}$. Divide the second number by the first and compare the quotient with one. For this, transform the obtained fraction by multiplying the numerator and the denominator by the same non-zero number. We get: $(\sqrt{2016+\sqrt{2015}}-\sqrt{2015+\sqrt{2016}}):(\sqrt{2016}-\sqrt{2015})=$ $=\frac{(2016+\sqrt{2015})-(2015+\sqrt{2016})}{(\sqrt{2016+\sqrt{2015}}+\sqrt{2015+\sqrt{2016}})(\sqrt{2016}-\sqrt{2015})}=$ $=\frac{(1+\sqrt{2015}-\sqrt{2016})(\sqrt{2016}+\sqrt{2015})}{\sqrt{2016+\sqrt{2015}}+\sqrt{2015+\sqrt{2016}}}=\frac{\sqrt{2016}+\sqrt{2015}-1}{\sqrt{2016+\sqrt{2015}}+\sqrt{2015+\sqrt{2016}}}\sqrt{2015}+\sqrt{2016+\sqrt{2015}}$

Grading criteria.

“+" - a complete and justified solution is provided

“士” - correct calculations are provided, but there are inaccuracies in justifications or logical errors “-” - only the answer is provided

“-" - the problem is not solved or solved incorrectly