text: # Posed in regional mathematics Olympiad 2005 [duplicate]
  
  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
  
  ## marked as duplicate by user10354138, Xander Henderson, max_zorn, Lord Shark the Unknown, NosratiNov 3 '18 at 7:12
  
  This question was marked as an exact duplicate of an existing question.
  
  • You should post your work and approach as well. – Arpan Mar 18 '15 at 15:19
  • Hint: The conclusion is equivalent to saying $\Delta<\frac{1}{16}$. – Thomas Andrews Mar 18 '15 at 15:20
  • @Thomas Andrews Yes, I have figured that out. – Aniket Bhattacharyea Mar 18 '15 at 15:22
  • If any value is $\leq 1/4$, you can show $\Delta<\frac{1}{16}$. So you can assume that $\frac{1}{4}< a,b,c<\frac{1}{2}$. – Thomas Andrews Mar 18 '15 at 15:24
  • @ThomasAndrews Is it okay to assume that? – Arpan Mar 18 '15 at 15:26
  
  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:
  
  $$\Delta\le\frac{\frac{1}{81}+\frac{2}{27}+\frac{1}{9}}{4}=\frac{4}{81} < \frac{1}{16}$$
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