Minimum value

By the Cauchy Schwarz Inequality,

$\displaystyle \biggl( \sum_{i=1}^n a_i ^2 \biggr) \biggl( \sum_{j=1}^n b_j^2 \biggr) \geq \biggl( \sum_{i=1}^n a_ib_i \biggr) ^2$

Let $a_i$ be $(\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d},\sqrt{e})$

Let $b_i$ be $\Bigl(\frac{2}{\sqrt{a}} , \frac{3}{\sqrt{b}},\frac{4}{\sqrt{c}},\frac{6}{\sqrt{d}} ,\frac{7}{\sqrt{e}}\Bigr)$

Now applying the inequality, we get

$\displaystyle (a+b+c+d+e)\biggl( \frac{4}{a}+\frac{9}{b} + \frac{16}{c} + \frac{36}{d}+\frac{49}{e} \biggr) \geq (2+3+4+6+7)^2$

$\displaystyle (a+b+c+d+e) \geq \dfrac{22^2}{8}$

$\displaystyle (a+b+c+d+e) \geq \dfrac{484}{8} = \boxed{60.5}$

Let $\dfrac{4}{a}+\dfrac{9}{b}+\dfrac{16}{c}+\dfrac{36}{d}+\dfrac{49}{e}=S$

By $AM \geq HM$

$\dfrac{2 \times\dfrac{a}{2}+3\times\dfrac{b}{3}+4\times\dfrac{c}{4}+6\times\dfrac{d}{6}+ 7\times\dfrac{e}{7}}{22} \geq \dfrac{22}{S}$

$a+b+c+d+e\geq \dfrac{22\times22}{8}$

$a+b+c+d+e\geq 60.5$

So $(a+b+c+d+e)_{min}=60.5$

by Titu's Lemma we have,

$\large{ \sum _{ i=1 }^{ n }{ \frac { \left( { x }_{ i }^{ 2 } \right) }{ \left( { a }_{ i } \right) } } \ge \frac { \left( \sum _{ i=1 }^{ n }{ ({ x }_{ i }) } \right) ^{ 2 } }{ \left( \sum _{ i=1 }^{ n }{ ({ a }_{ i }) } \right) } \\ \therefore \frac { { 2 }^{ 2 } }{ a } +\frac { { 3 }^{ 2 } }{ b } +\frac { { 4 }^{ 2 } }{ c } +\frac { { 6 }^{ 2 } }{ d } +\frac { { 7 }^{ 2 } }{ e } \ge \frac { (2+3+4+6+7)^{ 2 } }{ a+b+c+d+e } \\ \therefore \quad 8\quad \ge \frac { (22)^{ 2 } }{ a+b+c+d+e } \\ \therefore \quad a+b+c+d+e\ge \quad \frac { 484 }{ 8 } \\ i.e\quad a+b+c+d+e\ge \quad 60.5}\sum _{ i=1 }^{ n }{ \frac { \left( { x }_{ i }^{ 2 } \right) }{ \left( { a }_{ i } \right) } } \ge \frac { \left( \sum _{ i=1 }^{ n }{ ({ x }_{ i }) } \right) ^{ 2 } }{ \left( \sum _{ i=1 }^{ n }{ ({ a }_{ i }) } \right) } \\ \therefore \frac { { 2 }^{ 2 } }{ a } +\frac { { 3 }^{ 2 } }{ b } +\frac { { 4 }^{ 2 } }{ c } +\frac { { 6 }^{ 2 } }{ d } +\frac { { 7 }^{ 2 } }{ e } \ge \frac { (2+3+4+6+7)^{ 2 } }{ a+b+c+d+e } \\ \therefore \quad 8\quad \ge \frac { (22)^{ 2 } }{ a+b+c+d+e } \\ \therefore \quad a+b+c+d+e\ge \quad \frac { 484 }{ 8 } \\ i.e\quad a+b+c+d+e\ge \quad 60.5$