<s>[INST] 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.
 [/INST] Another idea to deal with this type of inequality.
Assume $c=\min\{a,b,c\} \to 0< c \le \dfrac{1}{3}.$
Let $$f(a,b,c)=2 \left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}+5\left(a^2+b^2+c^2\right)+30abc$$
$$g(a,b,c)=2 \left[ \frac{1}{2} \left( a+b \right) ^{2}+c^2 \right] ^2+\frac{5}{2}
 \left( a+b \right) ^{2}+5c^2+\dfrac{15}{2} \left( a+b \right) ^2c$$
We will so that
$$f(a,b,c)\ge g(a,b,c)\Leftrightarrow \left( a-b \right) ^{2} \left\{ 3a^2+2ab+3b^2+4{c}^{2}
+15\left(\frac{1}{3}-c\right) \right\}\ge 0\, \forall\,0<c\le \frac{1}{3}
$$
We only need to prove $$g(a,b,c)\ge 3\Leftrightarrow \frac{1}{2} c(c+1)(3c-1)^2\ge 0.$$
Done.
</s>