2
$\begingroup$

This question is an exact duplicate of:

Let a, b, c be three positive real numbers such that $a+b+c = 1$.

Let $\Delta= \min( a^{3} + a^{2}bc, b^{3}+ab^{2}c, c^{3}+abc^{2} )$.

Prove that the roots of the equation $x^{2} + x + 4 \Delta = 0$ are real. The last line is equivalent to $\Delta \leq \frac{1}{16} $. So I tried to prove by contradiction. I assumed all of them are $ > \frac{1}{16} $and tried to draw a contradiction with the fact $a+ b+ c=1$ but failed

$\endgroup$

marked as duplicate by user10354138, Xander Henderson, max_zorn, Lord Shark the Unknown, Nosrati Nov 3 '18 at 7:12

This question was marked as an exact duplicate of an existing question.

  • 1
    $\begingroup$ You should post your work and approach as well. $\endgroup$ – Arpan Mar 18 '15 at 15:19
  • $\begingroup$ Hint: The conclusion is equivalent to saying $\Delta<\frac{1}{16}$. $\endgroup$ – Thomas Andrews Mar 18 '15 at 15:20
  • $\begingroup$ @Thomas Andrews Yes, I have figured that out. $\endgroup$ – Aniket Bhattacharyea Mar 18 '15 at 15:22
  • $\begingroup$ If any value is $\leq 1/4$, you can show $\Delta<\frac{1}{16}$. So you can assume that $\frac{1}{4}< a,b,c<\frac{1}{2}$. $\endgroup$ – Thomas Andrews Mar 18 '15 at 15:24
  • $\begingroup$ @ThomasAndrews Is it okay to assume that? $\endgroup$ – Arpan Mar 18 '15 at 15:26
2
$\begingroup$

First, we see that the roots of the polynimium are real iff:

$$1-16\Delta \ge 0 \Leftrightarrow \Delta \le \frac{1}{16}$$

Assume without loss of gennerality that $a \le b \le c$. Then:

$$\Delta = a^3 + a^2bc = a^2(a+bc)$$

$bc$ is biggest when $b=c=\frac{1-a}{2}$, so:

$$\Delta\le a^2\left(a+\left(\frac{1-a}{2}\right)^2\right)=\frac{a^4+2a^3+a^2}{4}$$

Notice that this expression grows monotonically with $a$, but $a\le \frac13$, so it's maximum is at $a=\frac13$. But then:

$$\Delta\le\frac{\frac{1}{81}+\frac{2}{27}+\frac{1}{9}}{4}=\frac{4}{81} < \frac{1}{16}$$

$\endgroup$

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