TST

Let $p=a+b+c$, $q=ab+bc+ca$ and $r=abc$, we have $$2(p^2-2q)+3q=5p\text{ or }2p^2=5p+q.(1)$$

We need to prove that $4(p^2-2q)+2q+7r\le 25$ or $4p^2+7r\le 25+6q$. Since $q=p^2-5p$, the inequality is equivalent to: $$7r+30p\le 8p^2+25.$$ Notice that $(xy+yz+zx{)}^2\ge 3xyz(x+y+z)$ implies $q^2\ge 3pr$. We distinguish two cases regarding the value of $p$.

If $p=0$ then $a=b=c=0$, so the first inequality is true.

If $p>0$ then $r\le \frac{q^2}{3p}$, so we have to prove that $$\begin{array}{rl}& 7\frac{q^2}{3p}+30p\le 8p^2+25\\ & 7(2p^2-5p{)}^2+90p^2\le 24p^3+75p\\ & p(p-3)(2p-5)(14p-5)\le 0.\end{array}(2)$$ On the other hand, $2p^2-5p=q\le \frac{p^2}{3}$ implies $\frac{5}{2}\le p\le 3$, so inequality (2) is true. Therefore, the inequality (1) is true.