Prove that: $2a^2+3bc+4\ge b^2+c^2+5a$

Question

For $a,b,c\ge 0;a^2b+b^2c+c^2a=1;a=\max\{a,b,c\}.$ Prove that: $$2a^2+3bc+4\ge b^2+c^2+5a$$

I try to prove this inequality in the following way:

1. If $4-5a\ge 0 \rightarrow qed.$
2. If $4-5a<0 \rightarrow a>\dfrac{4}{5}.$ Let $$f(a)=2a^2+3bc+4-b^2-c^2-5a$$ $$f'(a)=4a-5<0$$ So we only prove this inequality for $a=a_{\min}.$

Okay, now notice that the equality holds when $a=1,b=1,c=0$ or $a=1,b=0,c=1.$ I try $$1=a^2b+b^2c+c^2a\le a^2\left(b+c\right)+c^2a \Rightarrow a \ge {\frac {\sqrt {c^4+4b+4c}-{c}^{2}}{2\left(b+c\right)}},$$ which implies one of the value of $a_{\min}.$ But now, the inequality seems be harder since the sqrt sign appears, and I stuck here.

Helps me!