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.
 A: 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.
A: Method 1let $p=a+b+c,q=ab+bc+ca,r=abc$
Now by schur $$a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)\ge 0$$ $$\iff p^4-5p^2q+4q^2+6pr\ge 0$$ Now we have to prove $$4 p^4 - 18 p^2 q + 8 q^2 + 30 p r\ge 0$$ or $$4(p^4-5p^2q+4q^2+6pr)+2(p^2q+3pr-4q^2)\ge 0$$ whcih is true because  $$p^2q+3pr-4q^2=a^3(b+c)+b^3(a+c)+c^3(a+b)-2a^2b^2-2b^2c^2-2c^2a^2\ge 0$$
Method 2 I see that i have clashed with an existing answer  so here is a different way
let $3u=a+b+c,3v^2=ab+bc+ca,w^3=abc$ then without even exanding we see that its lenear in $w^3$ Hence by uvw method it suffices to check when

*

*$c=0$
The inequality achieves the equality

*

*$a=b$ also let $t=\frac{c}{a}$ The inequality is $$ 2(2t^2+1)^2+5(2t^2+1)(2t+1)^2+30t^2(2t+1)-3(2t+1)^4\ge 0$$ which is nothing but$$4{(t-1)}^2(t+1)\ge 0$$
Method 3 This method is ugly  although it can be got by hand ,i used wolfram alpha to speed up my calculation
WLOG $a\ge b\ge c$ let $$f(a,b,c)=
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 
$$ $$f(a,b,c)-f(\frac{a+b}{2},\frac{a+b}{2},c)=1/2 (a - b)^2 (8 a^2 + 12 a b + 8 b^2 - 5 a c - 5 b c - 6 c^2)\ge 0$$
As inequality is homogenous WLOG   $c=1$  and $t=\frac{a+b}{2}$ we then have to prove $$f(t,t,1)\ge 0$$ $$\iff  2(2t^2+1)^2+5(2t^2+1)(2t+1)^2+30t^2(2t+1)-3(2t+1)^4\ge 0$$ which is nothing but $$4{(t-1)}^2(t+1)\ge 0$$
A: 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.
