3
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ Is it $\sqrt 3$ or $3 \sqrt 3$ on the right-hand side? $\endgroup$ – Martin R Nov 25 '19 at 20:26
  • 1
    $\begingroup$ 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. $\endgroup$ – WE Tutorial School Nov 26 '19 at 10:31
4
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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, $$\sum_{cyc}\sqrt{a^2+ab+b^2}\geq\sum_{cyc}\sqrt{\frac{3}{4}(a+b)^2}=3\sqrt3>\sqrt3.$$

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$ \frac{\sqrt{3}}{2}(a+b+c)= \frac{3\sqrt{3}}{2}>\sqrt{3} $$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.