Inequality $2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3$ for $a+b+c=1$ Let $a,b,c$ be positive numbers with $a+b+c=1$. Show that
$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc  \ge 3
$$
Equality holds for $a=b=c=\frac13$.  One could try homogenizing:
$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 \ge 0
$$
Expanding gives
$$
2 \sum_{cyc} a^4  -  \sum_{cyc} a^3(b+c) - 2  \sum_{cyc} a^2 b^2 + 2 a b c \sum_{cyc} a \ge 0
$$
but I didn't manage to advance further.

From the Schur inequality, we have
$$\sum a^2(a-b)(a-c) \geqslant 0,$$
equivalent to
$$a^4+b^4+c^4+abc(a+b+c) \geqslant \sum a^3(b+c). \qquad (1)$$
After expand, we write your inequality as
$$2\left[\sum a^4 + a b c(a+b+c)\right] \geqslant \sum a^3(b+c) + 2 \sum a^2 b^2.$$
Now, from $(1)$ we need to prove
$$\sum a^3(b+c) \geqslant 2 \sum a^2 b^2.$$
Which is true because by the AM-GM inequality, we have
$$\sum a^3(b+c) = \sum ab(a^2+b^2) \geqslant \sum ab \cdot 2ab = 2 \sum a^2 b^2.$$
Done.
