This is a classic problem, but I haven't seen it on the site before.
Suppose $x$ and $y$ are real numbers that both lie in the interval $(0,1)$. Prove $$x^y+y^x>1.$$
The following spoiler box contains a hint that indicates the solution I know. Does anyone know a more natural solution?
Use Bernoulli's inequality to show $$x^y\ge \frac{x}{x+y}.$$
Edit: I recently learned the source is the 1996 French math olympiad.
Well, let me point out a natural way to "guess" an auxiliary inequality you mentioned above. So we want to prove that $$x^y+y^x>1.$$ As in most problems of such type the the first idea to try is some convexity. It is highly unlikely that direct Jensen's would work so we want to think about weighted one, namely $$m_1f(x_1)+m_2f(x_2)\ge (m_1+m_2)f\left(\frac{m_1x_1+m_2x_2}{m_1+m_2}\right).$$ The simplest way to "create weights" is to separate $x$ and $y,$ in other words we rewrite out inequality in the form $$x^y+y^x=x\cdot \frac{1}{x^{1-y}}+y\cdot \frac{1}{y^{1-x}},$$ and this is also a natural way to rewrite it since we like positive powers $1-x$ and $1-y$ more than negative one. Now it is time to apply Jensen's and the simplest function one can think of is $f(t)=\frac{1}{t}.$ This leads to the inequality $$x^y+y^x\ge (x+y)f\left(\frac{x^{2-y}+y^{2-x}}{x+y}\right)=\frac{(x+y)^2}{x^{2-y}+y^{2-x}}.$$ Now we want to show that $(x+y)^2=x^2+xy+xy+y^2>x^{2-y}+y^{2-x}.$ So we would like to show that $x^2+xy\ge x^{2-y}$ and $y^2+xy\ge y^{2-x}.$ But the first inequality after cancelation reduces to $$x+y\ge x^{1-y},$$ or $$x^y\ge \frac{x}{x+y}.$$
