# if quadratic $x^2-(a+b+c)x+ab+bc+ca=0$ has complex roots, Then prove that $\sqrt{a}$,$\sqrt{b}$ and $\sqrt{c}$ are sides of triangle

The following quadratic equation $$x^2-(a+b+c)x+ab+bc+ca=0$$ has complex roots.

Prove that $\sqrt{a}$,$\sqrt{b}$ and $\sqrt{c}$ are sides-lengths of triangle, where $a,b,c \in \mathbb{R^+}$.

Since the quadratic has complex roots, we have Discriminant negative. So

$$(a+b+c)^2 \lt 4ab+4bc+4ca$$ $\implies$

$$a^2+b^2+c^2 \lt 2(ab+bc+ca)$$ $\implies$

$$c^2-2c(a+b)+(a-b)^2 \lt 0$$ $\implies$

$$(a+c-b)^2 \lt 4ac$$ $\implies$

we get

$$a+c-b \lt 2 \sqrt{ac}$$

any clue further?

From the given we obtain: $$\sum_{cyc}(2ab-a^2)>0$$ or $$(\sqrt{a}+\sqrt{b}+\sqrt{c})\prod_{cyc}(\sqrt{a}+\sqrt{b}-\sqrt{c})>0$$ because \begin{align} \sum_{cyc}(2ab-a^2)&=4ab-(a+b-c)^2\\ &=\left(2\sqrt{ab}\right)^2-(a+b-c)^2\\ &=(2\sqrt{ab}-a-b+c)(2\sqrt{ab}+a+b-c)\\ &=(c-(\sqrt{a}-\sqrt{b})^2)((\sqrt{a}+\sqrt{b})^2-c)\\ &=(\sqrt{c}-\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}+\sqrt{c}). \end{align} Let $a\geq b\geq c$.

Hence, $\sqrt{b}+\sqrt{c}-\sqrt{a}>0$ and rest multipliers are also positives.

Done!

• Typo in line 8: $+-$ should be just $+$. Also, do you need quite so many equality signs? – John Bentin Aug 24 '17 at 10:24
• @John Bentin I fixed. Thank you! We have no any equality because for example $\sqrt{a}+\sqrt{c}-\sqrt{b}>0$ automatically. – Michael Rozenberg Aug 24 '17 at 10:28
• I've reformatted to make it somewhat more readable. – John Hughes Aug 24 '17 at 10:30
• @John Hughes Thank you! – Michael Rozenberg Aug 24 '17 at 10:31
• The redundant equality signs were removed in John Hughes' reformatting. – John Bentin Aug 24 '17 at 10:33

You were on the right track. Continuing from where you left off, we can write your last inequality as $$(\surd c)^2-(\surd a-\surd b)^2>0.$$When we factorize this difference of squares, we see that $\surd c+\surd a-\surd b$ and $\surd c-\surd a+\surd b$ are either both positive or both negative. The latter can be ruled out, since their sum $2\surd c$ is positive. Hence $\surd c+\surd a>\surd b$ and $\surd c+\surd b>\surd a$. By the symmetry of the original data, $\surd a+\surd b>\surd c$ too.