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Algebra Difficulty 7.2 National olympiad, round 2 Prove it

35. For the cubic equation x3+ax2+bx+c=0x^{3} + a x^{2} + b x + c = 0 with three real roots, prove that a2−3b⩾0a^{2} - 3 b \geqslant 0, and a2−3b\sqrt{a^{2} - 3 b} is not greater than the difference between the largest and smallest roots (1988 US Mathematical Olympiad problem)

Solution

35. Let the three roots of the equation be p,q,rp, q, r, and p⩽q⩽rp \leqslant q \leqslant r. By the relationship between roots and coefficients, a=a= - (p+q+r),b=pq+qr+rp(p+q+r), b=p q+q r+r p. Thus,
a2−3b=(p+q+r)2−3(pq+qr+rp)=p2+q2+r2−(pq+qr+rp)=12[(p−q)2+(q−r)2+(r−p)2]⩾0\begin{aligned} a^{2}-3 b= & (p+q+r)^{2}-3(p q+q r+r p)= \\ & p^{2}+q^{2}+r^{2}-(p q+q r+r p)= \\ & \frac{1}{2}\left[(p-q)^{2}+(q-r)^{2}+(r-p)^{2}\right] \geqslant 0 \end{aligned}

On the other hand, a2−3b⩽r−p\sqrt{a^{2}-3 b} \leqslant r-p is equivalent to
a2−3b⩽(r−p)2⇔12[(p−q)2+(q−r)2+(r−p)2]⩽(r−p)2⇔(p−q)2+(q−r)2⩽(r−p)2\begin{aligned} a^{2}-3 b \leqslant(r-p)^{2} \Leftrightarrow & \frac{1}{2}\left[(p-q)^{2}+(q-r)^{2}+(r-p)^{2}\right] \leqslant(r-p)^{2} \Leftrightarrow \\ & (p-q)^{2}+(q-r)^{2} \leqslant(r-p)^{2} \end{aligned}

The last inequality is clearly true, because q−p⩾0,r−q⩾0q-p \geqslant 0, r-q \geqslant 0, we have
(r−p)2=[(r−q)+(q−p)]2=(p−q)2+(q−r)2+2(q−p)(r−q)⩾(p−q)2+(q−r)2\begin{aligned} (r-p)^{2}= & {[(r-q)+(q-p)]^{2}=(p-q)^{2}+(q-r)^{2}+2(q-p)(r-q) \geqslant } \\ & (p-q)^{2}+(q-r)^{2} \end{aligned}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.