Prove or disprove the inequality if $a,b,c>0$, $a \geq b+c$. Prove or disprove the inequality
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 7abc$$ if $$a,b,c>0, a \geq b+c.$$
I thought to use this evaluation:
$$a^2b+b^2c+c^2a \geq 3abc.$$
So we have:
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 3abc+3abc=6abc,$$ which is obvious that $$6abc<7abc.$$
Is it right? I'm embarrassed that in my solution I did not have to use the condition that $$a \geq b+c.$$
Any hint would help a lot. thanks!

Dividing by $abc > 0,$ the relation to be shown is $$  \frac{b+c}{a}  + \frac{c+a}{b} + \frac{a+b}{c} \overset?\ge 7$$
Set $a = b+c + d$  for $d \ge 0.$ We then need to argue that for $b,c>0, d\ge 0,$
$$ \frac{b+c}{b+c+d} + \frac{2c+d}{b} + \frac{2b+d}{c} \overset?\ge 5$$
The LHS is invariant under scaling, so wlog set $b = 1.$ We need to show that for $c > 0, d \ge 0,$ $$ J(c,d) := \frac{1 + c}{1 + c + d} + 2c + d + \frac{2 + d}{c} \overset?\ge 5.$$
For a fixed value of $c$, the derivative with respect to $d$ is $$ \partial_d J = 1 + \frac{1}{c} -\frac{(1+c)}{(1 + c + d)^2} \ge 1 + \frac{1}{c} - \frac{1}{1+c} > 0. $$ This means that for any fixed $c$, $J$ attains its minimum at $d = 0.$ So we need ot argue that $J(c,0) \ge 5$ for any $c> 0,$ i.e. that $$ 1 + 2c + \frac2c \overset?\ge 5 \iff c + \frac1c \overset?\ge 2.$$ But this holds via the AM-GM inequality. We conclude that the original inequality is true.
