# Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$

Good evening everyone, I want to prove the following:

Let $$a,b,c>0$$ be real numbers such that $$a+b+c=3$$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq \sqrt 3.$$

My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc} \sqrt{(a+b)^2-ab}$$

and now I want to apply Cauchy-Schwarz but it is the wrong direction.

• Is it $\sqrt 3$ or $3 \sqrt 3$ on the right-hand side? – Martin R Nov 25 '19 at 20:26
• Note that $$\sqrt{x^2+xy+y^2}=\sqrt{\frac{(2x+y)^2}{4}+\frac{3y^2}{4}}.$$ Then use the triangle inequality or Minkowski's inequality. – Batominovski Nov 26 '19 at 10:31

Lemma For all $$x$$ we have $$\sqrt{x^2+x+1}\geq {\sqrt{3}\over 2}(x+1)$$

Proof After squaring and clearing the denominator we get $$4x^2+4x+4\geq 3(x^2+2x+1)$$ which is the same as $$x^2-2x+1\geq 0$$

Using lemma we get $$\sqrt{a^2+ab+b^2}= b\sqrt{\Big({a\over b}\Big)^2+{a\over b}+1}\geq b\cdot {\sqrt{3}\over 2}({a\over b}+1)=$$ $$={\sqrt{3}\over 2}(a+b)$$ so

$$... \geq {\sqrt{3}\over 2}(a+b) + {\sqrt{3}\over 2}(b+c)+ {\sqrt{3}\over 2}(c+a) =3\sqrt{3}$$

By C-S $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}(a^2+ab+b^2+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}}=$$ $$=\sqrt{\sum_{cyc}(2a^2+ab++2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right)\left(\left(a+\frac{c}{2}\right)^2+\frac{3c^2}{4}\right)}}\geq$$ $$\geq\sqrt{\sum_{cyc}(2a^2+ab+2\left(\left(a+\frac{b}{2}\right)\left(a+\frac{c}{2}\right)+\frac{3bc}{4}\right)}=$$ $$=\sqrt{\sum_{cyc}(4a^2+5ab)}\geq\sqrt{\sum_{cyc}(3a^2+6ab)}=3\sqrt3>\sqrt3.$$ Also, $$\sum_{cyc}\sqrt{a^2+ab+b^2}\geq\sum_{cyc}\sqrt{\frac{3}{4}(a+b)^2}=3\sqrt3>\sqrt3.$$

Observe that, in virtue of the QM-AM inequality, we obtain $$\sqrt{a^2+ab+b^2}\geqslant \frac{a+\sqrt{ab}+b}{\sqrt3}$$ Similarly $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geqslant \frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}$$ We thus want to prove $$\frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}\geqslant\sqrt3\iff 6+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\geqslant 3$$ whicht is obsviously true

I’m not 100% sure but this is what I got.

$$a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2>(a+b/2)^2$$ From this we get $$a+b/2+b+c/2+c+a/2=9/2>\sqrt3$$ Or $$a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2> \frac{3}{4}b^2$$ From this we get $$\frac{\sqrt{3}}{2}(a+b+c)= \frac{3\sqrt{3}}{2}>\sqrt{3}$$

• Note that the RHS should be $3 \sqrt{3}$ instead. (OP had a typo in the question) – Calvin Lin Nov 27 '19 at 6:49
• Oh should i remove my answer? – user726500 Nov 27 '19 at 8:41