Let $a, b, c, d$ be such that $a+3 b+5 c+7 d=14$. Find the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}$ and specify the quadruplet $(a, b, c, d)$ for which this value is achieved.
The idea here is to use the Cauchy-Schwarz inequality to introduce a sum of squares, then get rid of the coefficients in front of \(a, b, c, d\) using calculations. To use Cauchy-Schwarz, we first need to square our equality. Therefore, \((a + 3b + 5c + 7d)^2 = 14^2 = 196\). And \(196 = (a + 3b + 5c + 7d)^2 \leq (a^2 + b^2 + c^2 + d^2)(1^2 + 3^2 + 5^2 + 7^2) = 84(a^2 + b^2 + c^2 + d^2)\), so \(a^2 + b^2 + c^2 + d^2 \geq \frac{196}{84} = \frac{7}{3}\). It remains to verify that this lower bound is achievable. The case of equality in the Cauchy-Schwarz inequality is achieved when \(\frac{a}{1} = \frac{b}{3} = \frac{c}{5} = \frac{d}{7}\). Since we have \(a + 3b + 5c + 7d = 14\), it follows that \(14 = a + 9a + 25a + 49a = 84a\), so \(a = \frac{14}{84} = \frac{1}{6}\) and we find that \(b = \frac{1}{2}\), \(c = \frac{5}{6}\), and \(d = \frac{7}{6}\). The quadruplet that achieves this minimum is therefore \(\left(\frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}\right)\). Conversely, we indeed have \(a + 3b + 5c + 7d = \frac{84}{6} = 14\) as well as \(a^2 + b^2 + c^2 + d^2 = \frac{84}{36} = \frac{7}{3}\), which concludes the proof.