EDITED

basketball9 wrote: $(b^3+ac^{\color{red}{2}})\left(\frac{1}{a}+\frac{1}{b}\right)\geq(b+c)^2$ $\Rightarrow b^3+ac^{\color{red}{2}} \geq\frac{ab(b+c)^2}{a+b}$ so ... I made in red the two necessary corrections. A rather more straightforward method. Divide by $a^3$ and denote $x=b/a$, $y=c/a$ (hence $xy>1$) to need to prove $x^3 + y^2 > x + xy$. Write this as $y^2 -xy + x^3 - x$, of discriminant $\Delta = x^2 - 4x^3 + 4x$. The discriminant is negative for $x > \frac {1 + \sqrt{65}} {8}$, and then the trinomial in $y$ is always positive. For $x \leq \frac {1 + \sqrt{65}} {8} < \sqrt{2}$ we compute that $\frac {x + \sqrt{\Delta}} {2} \leq \frac {1} {x}$, equivalent to $x\sqrt{\Delta} \leq 2-x^2$, and after squaring, $(x-1)^2(x+1)(x^2+x+1) \geq 0$ (equality occurs for $x=1$, when the trinomial is $y^2 - y >0$, since $y = xy > 1$). Thus for $0< y \leq \frac {x + \sqrt{\Delta}} {2}$ (the positive root of that trinomial) we have $y \leq \frac {1} {x}$, i.e. $xy \leq 1$, absurd; hence in fact we have $y$ larger than the root, so the trinomial takes positive vales, as needed.

homogeneous , let $a=1$ , we must proof $b^3 + c^2 > b(c+1)$. and we know $bc>1$ $b^3 + c^2 = b^3-b^2+b^2 +c^2 \ge b^3 - b^2 + 2bc > b^3 - b^2 + bc +1=b^3 - b^2 - b +1 +b +bc =$ $(b+1)(b-1)^2+b +bc \ge b+bc$ hence $b^3+c^2 >b + bc =b(c+1)$ and proof has finished.

Consider the graph of $y=f(x)=bx^2+2ax+c$, by $a>0$, the graph is convex and has the axis $x=-\frac{a}{c}$. Since $D/4=a^2-bc<0$ and $f(0)=c>0$, the graph is always above the $x$-axis and $f(x)$ has the minimum value $\frac{bc-a^2}{c}>0$ at $x=-\frac{a}{c}$, so we have $0<a<b<c,\ a^2-bc<0\Longleftrightarrow f(b)-f(a)=(b-a)(b^2+ab+2a)>0$ and $f(c)-f(b)=(c-a)(bc+ab+2a)>0$ $\Longleftrightarrow 0<a<b<c,\ a^2-bc<0\Longleftrightarrow f(a)<f(b)<f(c)$, thus $0<a<b<c\Longrightarrow b^3>a^2b$ and $ac^2>abc$, $\therefore b^3+ac^2>a^2b+abc$, yielding $b^3+ac^2>ab(a+c)$.

by AM_GM inequality ，see here http://www.om.edu.pl/index2.php?poziom=9&main=zadania.php

$f(a)=ba^2+a(bc-c^2)-b^3$ and $a\in [0,\sqrt{bc}]$ and$b>0$ So maxf(a)=f(0) or f($\sqrt{bc}$) and it's clear that $f(0)$ and $f(\sqrt{bc})$ is negative..

It's equivalent to $a(c-\frac{b}{2})^2+b^3-a^2b-\frac{ab^2}{4}>0$; it suffices to discuss two cases: $b\ge\sqrt2 a$ or $b\le\sqrt2 a$.

homogeneous, so let $abc=1$ , and $a=\frac{u}{v}$, $b=\frac{v}{w}$, $c=\frac{w}{u}$. Then we must to prove that: $\frac{v^3}{w^3}+\frac{w^2}{uv}>\frac{u^2}{vw}+1$. Or that: $uv^4+w^5>uvw^3+u^3w^2$. Remember, that $a^2<bc\rightarrow u<v$, ($u^3<v^3$) so if $w<u$ or $v<w$ than we can use rearrangement on $(uv;w^2)$ and $(v^3;w^3)$, else if $u<w<v$ than we have this: $uv^4>uvw^3$ and $w^5>u^3w^2$, which finishes proving.