If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$

Question

If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$

Normally when I do inequalities I try to first find where equality would be achieved, but in this case I have no idea where to start to find it.

From $a^2 - 2ab +b^2 \geq 0$ we can obviously get $a^2 -ab +b^2\geq 3ab$, and after substituting that, we can change the inequality to showing that $\sqrt{ab} + \sqrt{bc} +\sqrt{ac} \geq 1$, but clearly if a=b=c, then equality is not achieved. I wanted to try to use Cauchy-Schwarz or something of the sort, but those inequalities generally are used to find the upper bound of sums of roots.

Answer 1

It's true even for any reals $a$, $b$ and $c$ such that $a+b+c=3.$

Indeed, by C-S $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2\right)\left(\left(a+\frac{c}{2}\right)^2+\frac{3}{4}c^2\right)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(2a^2+ab+2\left(\left(a+\frac{b}{2}\right)\left(a+\frac{c}{2}\right)+\frac{3}{4}bc\right)\right)}=$$ $$=\sqrt{\sum_{cyc}(4a^2+5ab)}\geq\sqrt{3(a+b+c)^2}=\sqrt3|a+b+c|=3\sqrt3>\sqrt3.$$

For positive variables we have: $$\sum_{cyc}\sqrt{a^2+ab+b^2}\geq a+b+c=3>\sqrt3.$$

Answer 2

Use QM-AM to obtain \begin{align*}\sum_{cyc}\sqrt{a^2+ab+b^2}\geqslant \sum_{cyc}\frac{a+b+\sqrt{ab}}{\sqrt{3}}&=\frac{2(a+b+c)}{\sqrt{3}}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&=2\sqrt{3}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&>\sqrt{3}\end{align*}

Answer 3

By $a + b + c = 3$, the inequality is equivalent to $$ \sum_{cyc}{\sqrt{(3-c)^2-ab}}\geqslant \sqrt{3} $$ $$ \sum_{cyc}{9 - 6c + c^2 - ab}\geqslant 3 $$ Which is true because $\sum_{cyc}9-6c = 9$ and $a^2 + b^2 + c^2\geqslant ab + bc + ca$.

And $\sum_{cyc}{\sqrt{((3-c)^2-ab)((3-b)^2 - ca)}} \geqslant0$ is true.

Answer 4

$$\left(\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{12}=\frac{a^2+ab+b^2}{3}$$

$$\therefore\ \sqrt{\frac{a^2+ab+b^2}{3}}\ge\frac{a+b}{2}$$

$$\therefore\ \sum_{cyc}\sqrt{\frac{a^2+ab+b^2}{3}}\ge\sum_{cyc}\frac{a+b}{2}=a+b+c=3$$

So in fact a stronger identity is possible $\sum_{cyc}\sqrt{a^2+ab+b^2}\ge 3\sqrt{3}$.