If $a,b\in \mathbb R^+$, $ |a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}}$ Prove $a+b\leq 2$ 
Question:  If $a,b\in \mathbb R^+$, $|a-2b|\leq\frac {1}{\sqrt{a}}$,  $|b-2a|\leq\frac {1}{\sqrt{b}},$ prove $a+b≤2$.

I figured out that $a+b\leq \frac {1}{\sqrt{a}}+\frac {1}{\sqrt{b}}$, but I am not sure how to prove that $a+b\leq 2$ after doing this.
Can anyone help me?

Just my 2 cents to the Albus' answer, without using power mean. We can assume $0<a\leq b$ ($a=0$ can be checked easily) then $$\frac{(a+b)^2}{4}\leq a^3+b^3 \iff \frac{\left(1+\frac{b}{a}\right)^3}{4}\leq 1+\left(\frac{b}{a}\right)^3 \tag{1}$$
Now
$$f(x)=\frac{\left(1+x\right)^3}{4} -1-x^3=(1+x)\left(\frac{(1+x)^2}{4}-(1-x+x^2)\right)
=\\
(x+1)\left(\frac{-3+6x-3x^2}{4}\right)=-\frac{3}{4}(x+1)(x-1)^2$$
We can see that $f(x)\leq 0$ for $x\geq1$, which explains $(1)$.
