Minimum Fraction Product

By the form of Cauchy-Schwarz Inequality known as Titu's Lemma ,

$\displaystyle {(a+b+c+d+e) \cdot \left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+\frac{16}{d}+\frac{25}{e}\right)}$

$\displaystyle{\geq\left(\sqrt{a}\sqrt{\frac{1}{a}}+\sqrt{b}\sqrt{\frac{4}{b}}+\sqrt{c}\sqrt{\frac{9}{c}}+\sqrt{d}\sqrt{\frac{16}{d}}+\sqrt{e}\sqrt{\frac{25}{e}}\right)^2}$

$\geq(1+2+3+4+5)^2$

$\geq225$

Hence, the minimum value is $\boxed{225}$ .

The equality occurs where $\frac{1}{a} = \frac{2}{b} = \cdots = \frac{5}{e},$ so for example we can take $a=1,b=2,c=3,d=4,e=5.$

By Cauchy-Schwarz in Engel Form, we have:

$(a+b+c+d+e)(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+\frac{16}{d}+\frac{25}{d})\geq$

$(a+b+c+d+e)(\frac{(\sqrt{1}+\sqrt{4}+\sqrt{9}+\sqrt{16}+\sqrt{25})^{2}}{a+b+c+d+e})$

And $(1+2+3+4+5)^{2}=\boxed{225}$ .

By cauchy we have

$\large{\Rightarrow \quad \left\{ { { \sqrt { a } \sqrt { \frac { 1 }{ a } } +\sqrt { b } \sqrt { \frac { 4 }{ b } } +\sqrt { c } \sqrt { \frac { 9 }{ c } } +\sqrt { d } \sqrt { \frac { 16 }{ d } } +\sqrt { e } \sqrt { \frac { 25 }{ e } } } } \right\} ^{ 2 }\\ \le \left\{ { \sqrt { a } }^{ 2 }+{ \sqrt { b } }^{ 2 }+{ \sqrt { c } }^{ 2 }+{ \sqrt { d } }^{ 2 }+{ \sqrt { e } }^{ 2 } \right\} \left\{ \frac { 1 }{ a } +\frac { 4 }{ b } +\frac { 9 }{ c } +\frac { 16 }{ d } +\frac { 25 }{ e } \right\} \\ \therefore \quad \Rightarrow \quad (1+2+3+4+5)^{ 2 }\\ \le \left\{ a+b+c+d+e \right\} \left\{ \frac { 1 }{ a } +\frac { 4 }{ b } +\frac { 9 }{ c } +\frac { 16 }{ d } +\frac { 25 }{ e } \right\} \\ \\ \Rightarrow \quad \boxed{225}\le \left\{ a+b+c+d+e \right\} \left\{ \frac { 1 }{ a } +\frac { 4 }{ b } +\frac { 9 }{ c } +\frac { 16 }{ d } +\frac { 25 }{ e } \right\} }$

Cauchy-Schwarz Buniakowsky is full name of this inequality...