Quadratic Sum Minimization 1
Let $ a, $ $ b, $ $ c, $ $ d $ be real numbers such that $ a + b + c + d = 17 $ and $ ab + bc + cd + da = 46 $. Find the minimum possible value of $ a^2 + b^2 + c^2 + d^2 $.
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Solution
Note that $ ab + bc + cd + da = 46 $ factors as $ (a + c)(b + d) $. So, let $ r = a + c $ and $ s = b + d $. Then $ r + s = 17 $ and $ rs = 46, $ so by Vieta's formulas, $ r $ and $ s $ are the roots of $ x^2 - 17x + 46 = 0 $. Thus, $ r $ and $ s $ are equal to
\[\frac{17 \pm \sqrt{105}}{2},\]in some order.
We can let $ a = \frac{r}{2} + t, $ $ c = \frac{r}{2} - t, $ $ b = \frac{s}{2} + u, $ and $ d = \frac{s}{2} - u $. Then
\[a^2 + b^2 + c^2 + d^2 = \frac{r^2}{2} + 2t^2 +\frac{s^2}{2} + 2u^2 \ge \frac{r^2 + s^2}{2} = \frac{197}{2}.\]Equality occurs when $ a = c = \frac{r}{2} $ and $ b = d = \frac{s}{2}, $ so the minimum value is $ \boxed{\frac{197}{2}} $.
Freeflows