HINT

$$(a-b)^2(a^2-ab+b^2) \ge 0 \iff a^4+4a^2b^2+b^4 \ge 3ab(a^2+b^2) \iff a^5b+4a^3b^3+ab^5\ge 3a^4b^2+3a^2b^4 \iff (a^3+b^3)^2 \ge (a^2+b^2)^2(a^2-ab+b^2) \iff \frac {a^3+b^3}{\sqrt{a^2-ab+b^2}} \ge a^2+b^2 \implies \sum_{cyc} \frac {a^3+b^3}{\sqrt{a^2-ab+b^2}} \ge 2(a^2+b^2+c^2).$$Q.E.D

Devastator wrote: For all positive real numbers $a,b,c$, prove that $$\frac{a^3+b^3}{ \sqrt{a^2-ab+b^2} } + \frac{b^3+c^3}{ \sqrt{b^2-bc+c^2} } + \frac{c^3+a^3}{ \sqrt{c^2-ca+a^2} } \geq 2(a^2+b^2+c^2)$$ $$\sum_{cyc}\frac{a^3+b^3}{ \sqrt{a^2-ab+b^2} }=\sum_{cyc} \sqrt{(a+b)(a^3+b^3)}\geq\sum_{cyc}(a^2+b^2)=2\sum_{cyc} a^2$$

Easy one. Note that $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ thus, $$\frac{a^3+b^3}{\sqrt{a^2-ab+b^2}} =\sqrt{(a+b)(a^{3}+b^{3})}=\sqrt{a^{4}+b^{4}+ab(a^{2}+b^{2})} \geq \sqrt{a^{4}+b^{4}+2a^{2}b^{2}}=(a^{2}+b^{2})$$$$\implies \boxed{\sum_{cyc} \frac{a^3+b^3}{\sqrt{a^2-ab+b^2}} \geq 2(a^{2}+b^{2}+c^{2})}$$as desired. $\blacksquare$ $(\mathbb{QED})$