The cubic equation $x^3 + ax^2 + bx + c = 0$ has three real roots. Show that $a^2-3b\geq 0$, and that $\sqrt{a^2-3b}$ is less than or equal to the difference between the largest and smallest roots.
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Tags: AMC, USA(J)MO, USAMO, algebra, polynomial, Vieta, algebra unsolved
modularmarc101
27.07.2011 08:17
Moderator note: the original wording of this post was as follows: Quote: The cubic polynomial $ x^3 + ax^2 + bx + c $ has real coefficients and three real roots $ r \ge s \ge t $. Show that $ k = a^2 - 3b \ge 0 $ and that $ \sqrt {k} \le r - t $.
By Vieta's formulas,
$k = a^2 - 3b = (r+s+t)^2 - 3(rs+st+tr) = \frac{1}{2}(r-s)^2+\frac{1}{2}(s-t)^2+\frac{1}{2}(t-r)^2 \geq 0$,
with equality when $r=s=t$.
Also,
$r \geq s \geq t \implies (s-r)(s-t) \leq 0$
$\implies s^2 - rs - st + rt \leq 0$
$\implies (r^2 + s^2 + t^2) - (rs + st + tr) \leq r^2+t^2 - 2rt$
$\implies k \leq (r-t)^2$
$\implies \sqrt{k} \leq r-t$,
with equality when $s = r$ and/or $s=t$.
Wave-Particle
17.01.2018 07:12
By Vieta's, $a = -(r+s+t)$ and $b=rs+st+rt$. Hence, we wish to show $$(r+s+t)^2 - 3rs-3st-3rt \ge 0$$$$\implies r^2+s^2+t^2-rs-st-rt \ge 0.$$This is trivial by Cauchy.
Squaring both sides of the second inequality, it suffices to show that $a^2-3b = k \le (r-t)^2$. As calculated above, we wish to show $$r^2+s^2+t^2-rs-st-rt \le r^2-2rt+t^2$$Simplifying, $$s^2-rs-st+rt \le 0.$$We can now factor this to get $$(s-t)(s-r) \le 0.$$Clearly this is true since $r\ge s \ge t$.