Proof that if $x,y>0$ and $x+y=1$, then $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$

$$(2x)^{\frac 1 x}=\frac{1}{\left( \frac{1}{2x}\right) ^{\frac{1}{x}}}\leq \frac{1}{1+\frac{1}{x}\left( \frac{1}{2x}-1\right)}=\frac{2x^2}{2x^2-2x+1}$$ by Bernoulli’s inequality. The same holds for $y$ and one immediately computes that if $x+y=1$, $$\frac{2x^2}{2x^2-2x+1}+\frac{2y^2}{2y^2-2y+1}=2$$ and the result follows.(just observe that the denominator are the same $=x^2+y^2$)