For $a,b,c\ge0$, show $a^2 + b^2 + c^2 = 3$ implies $(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4)$.

Question

$a,b,c$ are nonnegative real numbers. If $a^2 + b^2 + c^2 = 3$, prove that $$(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4).$$

I tried using $(a^3 + b^3 + c^3)^2 >= 3(a^3 + b^3 + c^3)abc$ to make it maybe better.

I also tried matching the degree so that the inequality changes to $(a^3 + b^3 + c^3)^2 >= ((a^2 + b^2 + c^2)^3)/9 + 2/3 * (a^2 + b^2 + c^2)(a^4 + b^4 + c^4)$, but it made it more complicated. I also tried using QM>=AM in the version of higher degrees.

Answer 1

We need to prove that: $$(a^3+b^3+c^3)^2\geq\frac{(a^2+b^2+c^2)^3}{9}+\frac{2(a^4+b^4+c^4)(a^2+b^2+c^2)}{3}$$ or $$\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+18a^3b^3-2a^2b^2c^2)\geq0,$$ which is true by AM-GM: $$\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+18a^3b^3-2a^2b^2c^2)\geq$$ $$\geq\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+16a^3b^3)=$$ $$=\sum_{cyc}(a^6-9a^4b^2+16a^3b^3-9a^2b^4+b^6)=\sum_{cyc}(a-b)^4(a^2+4ab+b^2)\geq0.$$