Taken from here in German. Solutions can be found in this link.

Attachments:
finalrunde_musterloesung_2018.pdf (269kb)
finalrunde_2018_de.pdf (273kb)

Amir Hossein wrote: Let $a,b,c,d,e$ be positive real numbers. Find the largest possible value for the expression $$\frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2}.$$ Set $k=\sqrt{\frac83}$ and \[P = 2a^2+b^2+2c^2+d^2+2e^2-k(ab+bc+cd+de).\]We have \[P = \frac{(4a-bk)^2+(bk+dk-4c)^2+(dk-4e)^2}{8} + \frac{(2bk^2+dk^2-8b)^2}{16(4-k^2)} \geqslant 0.\]Therefore \[\frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2} \leqslant \frac1k = \sqrt{\frac38}.\]Equality occur when $a = e, b =d = \sqrt6e, c = 2e.$

Amir Hossein wrote: Let $a,b,c,d,e$ be positive real numbers. Find the largest possible value for the expression $$\frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2}.$$ Solution (Arnaud): By AM-GM, $$2a^2+\frac{1}{3}b^2\geq\sqrt{\frac{8}{3}}ab , \frac{2}{3}b^2+c^2\geq\sqrt{\frac{8}{3}}bc , c^2+\frac{2}{3}d^2\geq\sqrt{\frac{8}{3}}cd , \frac{1}{3}d^2+2e^2\geq\sqrt{\frac{8}{3}}de ,$$$$2a^2+b^2+2c^2+d^2+2e^2\geq\sqrt{\frac{8}{3}}(ab+bc+cd+de) , $$$$\frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2}\leq\sqrt{\frac{3}{8}}, ...$$ Switzerland Mathematical Olympiad

Amir Hossein wrote: Let $a,b,c,d,e$ be positive real numbers. Find the largest possible value for the expression $$\frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2}.$$ Note that: $$2a^2+\alpha b^2\geq2\sqrt{2\alpha}ab,$$$$(1-\alpha)b^2+\beta c^2\geq2\sqrt{(1-\alpha)\beta}bc$$$$(2-\beta)c^2+\gamma d^2\geq2\sqrt{(2-\beta)\gamma}cd$$and $$(1-\gamma)d^2+2e^2\geq2\sqrt{2(1-\gamma)}de,$$where $0\leq \alpha\leq1$, $0\leq\beta\leq2$, $0\leq\gamma\leq1$ and $$2\alpha=(1-\alpha)\beta=(2-\beta)\gamma=2(1-\gamma),$$which gives $\alpha=\frac{1}{3}.$ Easy to see that the equality occurs, which gives a largest value: $$\frac{1}{2\sqrt{2\cdot\frac{1}{3}}}=\sqrt{\frac{3}{8}}.$$

you assumed this right ?? $$2\alpha=(1-\alpha)\beta=(2-\beta)\gamma=2(1-\gamma),$$

Yes, because I want that all coefficients before $ab$, $bc$, $cd$ and $de$ would be equal.