Let $x=a+b, y=\sqrt{ab}$; then $x\ge 2y > 0$. $\sqrt{ab} \leq \dfrac{1}{3} \sqrt{\dfrac{a^2+b^2}{2}}+\dfrac{2}{3} \dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}\Leftrightarrow 3y\le\sqrt{\frac{x^2-2y^2}{2}}+\frac{4y^2}{x}$ $\Leftrightarrow(x^2-2y^2)x^2\ge 2(3xy-4y^2)^2\Leftrightarrow (x-2y)^2[x^2+4y(x-2y)]\ge 0.$

In other words, $GM \leq \dfrac{1}{3} QM + \dfrac{2}{3} HM$, where $GM$, $QM$, $HM$ are the $\underline{\textrm{G}}$eometric, $\underline{\textrm{Q}}$uadratic, respectively $\underline{\textrm{H}}$armonic means of $a$ and $b$. Writing the inequality as $QM - GM \geq 2(GM - HM)$, and with the substitution $b=x^2a$ ($x>0$), the inequality writes in turn in the equivalent forms \[\sqrt{\dfrac {1+x^4} {2}} - x \geq 2\left (x - \dfrac {2x^2} {1+x^2} \right ) = \dfrac {2x(1-x)^2} {1+x^2},\] \[\dfrac {\dfrac {1+x^4} {2} - x^2} {\sqrt{\dfrac {1+x^4} {2}} + x} \geq \dfrac {2x(1-x)^2} {1+x^2},\] \[\dfrac {(1 - x^2)^2} {\sqrt{\dfrac {1+x^4} {2}} + x} \geq \dfrac {4x(1-x)^2} {1+x^2}.\] Factorizing $(1-x)^2$, we may continue with \[(1 + x)^2(1+x^2) \geq 4x\left (\sqrt{\dfrac {1+x^4} {2}} + x \right ),\] \[1 + 2x - 2x^2 + 2x^3 + x^4 \geq 4x\sqrt{\dfrac {1+x^4} {2}},\] \[\left ( (1 + x^4) - 4x\sqrt{\dfrac {1+x^4} {2}} + 2x^2\right ) + \left (2x - 4x^2 + 2x^3 \right ) \geq 0,\] \[2\left ( \sqrt{\dfrac {1+x^4} {2}} - x\right )^2 + 2x\left ( 1-x \right )^2 \geq 0.\] This last form being a sum of squares, it holds true. Equality obviously occurs if and only if $x=1$, that is, for $a=b$. I am not that happy with this solution of mine, which in fact does not capture the essence of the problem, being nothing more than a simple verification. I wait for some more "insightful" proof; also, I ask myself in which way (if any) can this be extended to $n> 2$ positive variables, given the classical inequalities \[\sqrt{\dfrac {\sum\limits_{k=1}^n x_k^2} {n}} \geq \dfrac {\sum\limits_{k=1}^n x_k} {n} \geq \sqrt[n]{\prod\limits_{k=1}^n x_k} \geq \dfrac {n} {\sum\limits_{k=1}^n \dfrac {1} {x_k}}.\]

it is easy to see inequality homogeneous. so we can take $ab=1$ and $a+b=x $ then $\leftrightarrow$ \[1 \leq \frac{1}{3}\sqrt{\frac{x^{2}-1}{2}}+\frac{4}{3x} \leftrightarrow (x-2)^{2}(x^{2}+4x-8)\geq 0\] it is true,because $x=a+b\geq 2\sqrt{ab}=2$

we also can take $a+b=2, \sqrt{ab}=x$; then $0<x\le 1$. $\sqrt{ab} \leq \dfrac{1}{3} \sqrt{\dfrac{a^2+b^2}{2}}+\dfrac{2}{3} \dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}\Leftrightarrow \sqrt{2-x}+2x\ge 3\sqrt{x}.$ here