THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

If $a>1$, taking logs of both sides we obtain the equivalent inequality $$\sin x+\cos x\cdot {\log }_a(a+1)\ge 1.$$ For $x\in [0,\pi /2]$, we have $\sin x\ge {\sin }^{2}x$ and $\cos x\ge {\cos }^{2}x$. Because ${\log }_a(a+1)>1$, we obtain $$\sin x+\cos x\cdot {\log }_a(a+1)\ge {\sin }^{2}x+{\cos }^{2}x=1.$$ The inequality clearly holds true for $a=1$.

Finally, if $a\in (0,1)$, the inequality is equivalent to $$\sin x+\cos x\cdot {\log }_a(a+1)\le 1,\forall x\in [0,\pi /2],$$ which is obvious, since $\sin x\le 1$, $\cos x\ge 0$ and ${\log }_a(a+1)<0$.