50 Followers Problem!

By Cauchy-Schwarz Inequality we have

$\large{{ \left( { a }^{ 2 }+{ c }^{ 2 } \right) \left( { b }^{ 2 }+{ d }^{ 2 } \right) }\ge { \left( ab+cd \right) }^{ 2 }}$

We will go back to the basics of Chebysev's Identities which stated that if $a_{1}\ge a_{2} \ge a_{3}........a_{n-1} \ge a_{n}$ and $b_{1} \ge b_{2} \ge b_{3}.........b_{n-1}\ge b_{n}$ .

$\large{{ \frac { \sum _{ i=1 }^{ n }{ \left( { a }_{ i } \right) \left( { b }_{ i } \right) } }{ n } }\ge \frac { \prod _{ i=1 }^{ n }{ { a }_{ i } } }{ n } \times \frac { \prod _{ i=1 }^{ n }{ { b }_{ i } } }{ n } \ge \frac { \sum _{ i=1 }^{ n }{ \left( { a }_{ i } \right) \left( { b }_{ n+1-i } \right) } }{ n } }$

In the question we are given $a \ge d, b\ge c$ . Applying Cheybsev's Identity

$\large{\left( ab+cd \right) \ge \left( ac+bd \right) \\ \left( ab+cd \right) \ge 7\\ { \left( ab+cd \right) }^{ 2 }\ge 49}$

And from the first equation we have

$\large{{ \left( { a }^{ 2 }+{ c }^{ 2 } \right) \left( { b }^{ 2 }+{ d }^{ 2 } \right) }\ge { \left( ab+cd \right) }^{ 2 }\\ \left( { a }^{ 2 }+{ c }^{ 2 } \right) \left( { b }^{ 2 }+{ d }^{ 2 } \right) \ge 49\\ \left( { b }^{ 2 }+{ d }^{ 2 } \right) \ge \frac { 49 }{ 9 } =5.44}$

equality holds when $\large{a=\dfrac{27}{\sqrt{130}}, b=c=\dfrac{21}{\sqrt{130}}, d=\dfrac{49}{3\sqrt{130}} }$