Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$ Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$
My proof :D
We have the inequality
$\sum_{cyc}^{ }a^2.\sum_{cyc}^{ }a\geq \sum_{cyc}^{ }a^2b $
which is equivalent to $\sum_{cyc}^{ }b(a-b)^2$ (true)(1)
In another side, by AM-GM, we have: $\sum_{cyc}^{ }a\leq \sum_{cyc}^{ }\dfrac{a^4+1+1+1}{4}=3$ (2)
Thus, from (1) and (2), we've completed our proof.
 A: Let $0\le z \le 3$. Then, we have the inequality $z^2 - 2z - 3 \le 0$. We need this because we have $\frac{a^2+b^2+c^2}{3} \le \left(\frac{a^4 + b^4 + c^4}{3}\right)^{1/2} = 1$. 
Now, $a^2b+b^2c+c^2a \le (a^2+b^2+c^2)^{1/2}(a^2b^2+b^2c^2+c^2a^2)^{1/2}$, thus we need to prove $a^2b^2+b^2c^2+c^2a^2 \le a^2+b^2+c^2$. We have
$$2(a^2b^2+b^2c^2+c^2a^2) = (a^2+b^2+c^2)^2 - (a^4+b^4+c^4) = (a^2+b^2+c^2)^2 -3.$$
Thus, if we define $a^2+b^2+c^2 = z$, we need to prove $z^2-2z-3 \le 0$, which is what we started with.
A: It's enough to prove our inequality for non-negative variables.
We'll prove a stronger inequality:
$$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3,$$
which solves our problem.
Indeed, by C-S $$a^2+b^2+c^2=\sqrt{(a^2+b^2+c^2)^2}\leq\sqrt{(1+1+1)(a^4+b^4+c^4)}=3.$$
Thus, 
$$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3=(a^2+b^2+c^2)\sqrt{\frac{a^2+b^2+c^2}{3}}\leq a^2+b^2+c^2.$$
Since the inequality $$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3$$
is homogeneous, we can assume that $a^2+b^2+c^2=3$ and we need tho prove that:
$$a^2b+b^2c+c^2a\leq3.$$
Now, let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, be Rearrangement and AM-GM we obtain:
$$a^2b+b^2c+c^2a=a\cdot ab+b\cdot bc+c\cdot ca\leq x\cdot xy+y\cdot xz+z\cdot yz=$$
$$=y(x^2+xz+z^2)\leq y\left(x^2+z^2+\frac{x^2+z^2}{2}\right)=\frac{3}{2}y(3-y^2)\leq3$$ and we are done!
