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1973 IMO Problems/Problem 3

Revision as of 00:50, 28 November 2017 by Myl (talk | contribs) (→‎Solution)

Let $a$ and $b$ be real numbers for which the equation $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution. For all such pairs $(a, b)$, find the minimum value of $a^2 + b^2$.

Solution

Substitute $z=x+1/x$ to change the original equation into $z^2+az+b-2=0$. This equation has solutions $z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}$. We also know that $|z|=|x+1/x| \geq 2$. So,

$\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2$

$\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2$

$|a|+\sqrt{a^2+8-4b} \geq 4$

Rearranging and squaring both sides,

$a^2+8-4b \geq a^2-16|a|+16$

$2|a|-b \geq 2$

So, $a^2+b^2 \geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\frac{4}{5})^2+\frac{4}{5}$.

Therefore, the smallest possible value of $a^2+b^2$ is $\frac{4}{5}$, when $a=\pm \frac{4}{5}$ and $b=\frac{-2}{5}$.

Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln733.html

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