66th Belarusian Mathematical Olympiad

By condition, we have $$a+\sin b>c+\sin d,(1)$$ $$\sin a+b>\sin c+d.(2)$$ Suppose, contrary to our claim, that $$c+d\ge a+b.\text{\hspace{1em}(3)}$$ Summing (1) and (3), we get $$a+\sin b+c+d>c+\sin d+a+b⟹d-\sin d>b-\sin b.(4)$$ Summing (2) and (3), we get $$\sin a+b+c+d>\sin c+d+a+b⟹c-\sin c>a-\sin a.(5)$$ Let $f(x)=x-\sin x$. Since $f^{\prime }(x)=1-\cos x\ge 0$ and there are no intervals such that $f^{\prime }(x)=0$, we see that the function $f$ is strictly increasing. By the same argument, the function $g(x)=x+\sin x$ is also strictly increasing. Therefore, from (4) and (5) we have $$d>b\Rightarrow d+\sin d>b+\sin b,(6)$$ and $$c>a⟹c+\sin c>a+\sin a.\text{\hspace{1em}(7)}$$ Summing (1) and (6), we get $$a+\sin b+d+\sin d>c+\sin d+b+\sin b\Rightarrow a+d>b+c,$$ and summing (2) and (7), we get $$\sin a+b+c+\sin c>\sin c+d+a+\sin a⟹b+c>a+d.$$ This contradiction proves the required statement.