Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$

Question

Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$

So, using AM-GM, or just pop out squares under square roots we can show: $$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}),$$ i.e. we need next to show that $(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge (a + b + c)$, but i don't know how to do it.

Any help appreciated

Answer 1

By Minkowski (triangle inequality) $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sum_{cyc}\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2}\geq$$ $$\geq\sqrt{\left(\sum_{cyc}\left(a+\frac{b}{2}\right)\right)^2+\frac{3}{4}\left(\sum_{cyc}b\right)^2}=\sqrt3(a+b+c).$$

Answer 2

Consider a point $O$ and the lines $OA$, $OB$, $OC$ of lengths $a,b,c$ respectively such that $\angle AOB=\angle BOC= \angle COA=120^°$. Consider the triangle $\Delta ABC$. using cosine rule you can see that the side lengths of that triangle are $\sqrt{a^2+ab+b^2},\sqrt{b^2+bc+c^2},\sqrt{c^2+ca+a^2}$.

By definition $O$ is the Fermat point of $\Delta ABC$. See the properties of Fermat point and solve this geometric inequality.

Answer 3

Also, we can use SOS here:

$$\sum_{cyc}\sqrt{a^2+ab+b^2}-\sqrt3(a+b+c)=\sum_{cyc}\left(\sqrt{a^2+ab+b^2}-\frac{\sqrt3(a+b)}{2}\right)=$$ $$=\sum_{cyc}\frac{4(a^2+ab+b^2)-3(a+b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}=\sum_{cyc}\frac{(a-b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}\geq0.$$

Answer 4

The problem has appeared here various times and maybe elsewhere too for sure. The key to this is to observe the relationship between: $(a-b)^2 \ge 0$ and $a^2+ab+b^2 \ge \dfrac{3(a+b)^2}{4}$. They are equivalent. Using the latter but taking the square root first $3$ times for $(a,b), (b,c),(c,a)$ and add up, yielding the result.