2016/022) Show that $a^b+b^a > 1$ for a , b positive

if a or b are above 1  we are done,.
so let us assume both $< 1$
by Bernoulli Inequality  we have
$(1+x)^n \>= 1+nx$
so $(1+\frac{a}{b})^{\frac{1}{a}}  \ge 1+\frac{1}{b}$
so $(1+\frac{a}{b})^{\frac{1}{a}} > \frac{1}{b}$
or $1+\frac{a}{b} > (\frac{1}{b})^a$
or $\frac{a+b}{b} > (\frac{1}{b})^a$
or $\frac{b}{a+b} < b^a$
similarly
$\frac{a}{a+b} < a^b$
Adding, we get $1 < a^b + b^a$
or $a^b+b^a > 1$