11. (SWE 4) ${ }^{\mathrm{IMO} 3}$ Determine the minimum of $a^{2}+b^{2}$ if $a$ and $b$ are real numbers for which the equation 
$$ x^{4}+a x^{3}+b x^{2}+a x+1=0 $$
has at least one real solution.
11. Putting $x+\frac{1}{x}=t$ we also get $x^{2}+\frac{1}{x^{2}}=t^{2}-2$, and the given equation reduces to $t^{2}+a t+b-2=0$. Since $x=\frac{t \pm \sqrt{t^{2}-4}}{2}, x$ will be real if and only if $|t| \geq 2, t \in \mathbb{R}$. Thus we need the minimum value of $a^{2}+b^{2}$ under the condition $a t+b=-\left(t^{2}-2\right),|t| \geq 2$. However, by the Cauchy-Schwarz inequality we have 
$$ \left(a^{2}+b^{2}\right)\left(t^{2}+1\right) \geq(a t+b)^{2}=\left(t^{2}-2\right)^{2} $$
It follows that $a^{2}+b^{2} \geq h(t)=\frac{\left(t^{2}-2\right)^{2}}{t^{2}+1}$. Since $h(t)=\left(t^{2}+1\right)+\frac{9}{t^{2}+1}-6$ is increasing for $t \geq 2$, we conclude that $a^{2}+b^{2} \geq h(2)=\frac{4}{5}$. The cases of equality are easy to examine: These are $a= \pm \frac{4}{5}$ and $b=-\frac{2}{5}$. Second solution. In fact, there was no need for considering $x=t+1 / t$. By the Cauchy-Schwarz inequality we have $\left(a^{2}+2 b^{2}+a^{2}\right)\left(x^{6}+x^{4} / 2+x^{2}\right) \geq$ 
$\left(a x^{3}+b x^{2}+a x\right)^{2}=\left(x^{4}+1\right)^{2}$. Hence 
$$ a^{2}+b^{2} \geq \frac{\left(x^{4}+1\right)^{2}}{2 x^{6}+x^{4}+2 x^{2}} \geq \frac{4}{5} $$
with equality for $x=1$.