We use the common notations: $a,b,c\,$ the side lengths; $\displaystyle s=\frac{a+b+c}{2}\,$ the semiperimeter; $[\Delta ABC]=S\,$ the area of $\Delta ABC.$

Let $s-a=x,\,$ $s-b=y,\,$ $s-c=z.\,$ Then

$\displaystyle \frac{1}{r_b}=\frac{y}{S},\,$ $\displaystyle \frac{1}{r_c}=\frac{z}{S},\,$ $\displaystyle \frac{1}{h_a}=\frac{y+z}{2S}.$

Denote $\displaystyle \frac{y}{S}=u\,$ and $\displaystyle \frac{z}{S}=v.\,$ The required inequality reduces to

$\sqrt{u^2+v+1}+\sqrt{v^2+u+1}\ge 2\sqrt{(u+v)^2+2(u+v)+4},$

which, after squaring becomes

$2\sqrt{(u^2+v+1)(v^2+u+1)}\ge 2uv+u+v+2.$

Let $u+v=2m\,$ and $uv=n.\,$ Squaring again leads to $(m^2-n)(8m+3)\ge 0,\,$ which is true, because, by the AM-GM inequality, $m^2\ge n.\,$

I am grateful to Leo Giugiuc who kindly communicated to me a problem by Nguyen Viet Hung, along with a solution of his. The problem was originally posted at the mathematical inequalities facebook group.