If $a,b,c\ge 0$ such that $ab+bc+ca=3$, then

$$4(a^3+b^3+c^3)+7abc+125\ge 48(a+b+c).$$

Proof(Vo Quoc Ba Can)

The inequality is equivalent to \[4(a^3+b^3+c^3)+7abc +125 \ge 16(ab+bc+ca)(a+b+c),\]

or

\[125 \ge A,\]

where $A=16(a+b+c)(ab+bc+ca)-4(a^3+b^3+c^3) -7abc.$

From the equivalence, it is obvious that the inequality holds for $A \le 0,$ so we only have to consider it when $A >0.$ In that case, it is equivalent to \[125^2 \ge A^2,\] and there are two possible cases can occur (assume that $a \ge b \ge c$):
1、Case $6(a+c) \ge 7b.$

Using the AM-GM inequality, we have

\[A^2 =(2a+2c+b)^2\cdot \frac{A}{2a+2c+b}\cdot \frac{A}{2a+2c+b} \le \left[\frac{(2a+2c+b)^2+2\cdot \frac{A}{2a+2c+b}}{3}\right]^3.\]

Thus, it suffices to prove that

\[25\cdot 3 \ge (2a+2c+b)^2+\frac{2A}{2a+2c+b},\]

or

\[25(ab+bc+ca) \ge (2a+2c+b)^2+\frac{2A}{2a+2c+b}.\]

An easy computation shows that, for the last inequality, we have

\[LHS-RHS =\frac{(a-b)(b-c)(6a+6c-7b)}{2a+2c+b},\]

which is obviously nonnegative.
2、Case $6(a+c) \le 7b.$

Applying the AM-GM inequality in the same way as the first case, but this time we change the positions of $a$ and $b$ ($a\to b$ and $b\to a$), it leads us to the inequality

\[25(ab+bc+ca) \ge (2b+2c+a)^2+\frac{2A}{2b+2c+a},\]

which is equivalent to

\[\frac{(a-b)(a-c)(7a-6b-6c)}{2b+2c+a} \ge 0.\]

This is true because

$$(a-b)(a-c) \ge 0$$

and

\[7a-6b-6c=13a-6(a+b+c) \ge 13b-6(a+b+c)=7b-(6a+6c) \ge 0.\]
The proof is completed. $\blacksquare$