Mathematical Enigma - XX

Using Titu's Lemma ,

$\dfrac { { 1 }^{ 2 } }{ a } +\dfrac { { 2 }^{ 2 } }{ b } +\dfrac { { 3 }^{ 2 } }{ c } +\dfrac { { 4 }^{ 2 } }{ d } +\dfrac { { 5 }^{ 2 } }{ e } +\dfrac { { 6 }^{ 2 } }{ f } \ge \dfrac { { (1+2+3+4+5+6) }^{ 2 } }{ a+b+c+d+e+f } =\dfrac { 21\times 21 }{ 7 } =63$ .

Use the concept $Weighted \; A.M \geq Weighted \; H.M$ , to get :

$\displaystyle \frac { 1(a)+2(\frac { b }{ 2 } )+3(\frac { c }{ 3 } )+4(\frac { d }{ 4 } )+5(\frac { e }{ 5 } )+6(\frac { f }{ 6 } ) }{ 21 } \ge \frac { 21 }{ 1(\frac { 1 }{ a } )+2(\frac { 2 }{ b } )+3(\frac { 3 }{ c } )+4(\frac { 4 }{ d } )+5(\frac { 5 }{ e } )+6(\frac { 6 }{ f } ) }$

$\displaystyle \Rightarrow \frac { a+b+c+d+e+f }{ 21 } \ge \frac { 21 }{ \frac { 1 }{ a } +\frac { 4 }{ b } +\frac { 9 }{ c } +\frac { 16 }{ d } +\frac { 25 }{ e } +\frac { 36 }{ f } }$

$\Rightarrow \displaystyle \frac { 1 }{ a } +\frac { 4 }{ b } +\frac { 9 }{ c } +\frac { 16 }{ d } +\frac { 25 }{ e } +\frac { 36 }{ f } \ge 63$

Let the given expression be equal to $x$

We will use Power Mean (QAGH) .

Applying $A.M.\ge H.M.$ on the numbers $a,\dfrac{b}{2},\dfrac{b}{2},\dfrac{c}{3},\dfrac{c}{3},\dfrac{c}{3},\dfrac{d}{4},\dfrac{d}{4},\dfrac{d}{4},\dfrac{d}{4},\dfrac{e}{5},\dfrac{e}{5},\dfrac{e}{5},\dfrac{e}{5},\dfrac{e}{5},\dfrac{f}{6},\dfrac{f}{6},\dfrac{f}{6},\dfrac{f}{6},\dfrac{f}{6},\dfrac{f}{6}$

$\dfrac{7}{21} \ge \dfrac{21}{x}$

$x \ge 63$

Define in $R^n$ Let $\vec{u} = (\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d},\sqrt{e},\sqrt{f}) \\ \vec{v} = (\frac{1}{\sqrt{a}},\frac{2}{\sqrt{b}},\frac{3}{\sqrt{c}},\frac{4}{\sqrt{d}},\frac{5}{\sqrt{e}},\frac{6}{\sqrt{f}})$

By Cauchy-Schwarz Inequality:

$\|\vec{u}\| \cdot \|\vec{v}\| \geq \vec{u} \cdot \vec{v} \\ \sqrt{(\sqrt{a})^{2} + (\sqrt{b})^{2} + (\sqrt{c})^{2} + (\sqrt{d})^{2} + (\sqrt{e})^{2} + (\sqrt{f})^{2}} \sqrt{(\frac{1}{\sqrt{a}})^{2} + (\frac{2}{\sqrt{b}})^{2} + (\frac{3}{\sqrt{c}})^{2} + (\frac{4}{\sqrt{d}})^{2} + (\frac{5}{\sqrt{e}})^{2} + (\frac{6}{\sqrt{f}})^{2}} \geq (\sqrt{a})(\frac{1}{\sqrt{a}}) + (\sqrt{b})(\frac{2}{\sqrt{b}}) + (\sqrt{c})(\frac{3}{\sqrt{c}}) + (\sqrt{d})(\frac{4}{\sqrt{d}}) + (\sqrt{e})(\frac{5}{\sqrt{e}}) + (\sqrt{f})(\frac{6}{\sqrt{f}}) \\ \sqrt{a+b+c+d+e+f} \sqrt{\frac{1}{a} + \frac{4}{b} + \frac{9}{c} + \frac{16}{d} + \frac{25}{e} + \frac{36}{f}} \geq 1+2+3+4+5+6 \\ \sqrt{7} \sqrt{\frac{1}{a} + \frac{4}{b} + \frac{9}{c} + \frac{16}{d} + \frac{25}{e} + \frac{36}{f}} \geq 21 \\ \frac{1}{a} + \frac{4}{b} + \frac{9}{c} + \frac{16}{d} + \frac{25}{e} + \frac{36}{f} \geq \frac{21^2}{7} = 63$

Equality holds when

$\frac{a}{1} = \frac{b}{2} = \frac{c}{3} = \frac{d}{4} = \frac{e}{5} = \frac{f}{6} = k, k \epsilon R \\ a = k, b = 2k, c = 3k, d = 4k, e = 5k, f = 6k \\ a + b + c + d + e + f = 21k = 7 \Rightarrow k = 1/3$

Therefore the minimum value is 63, and it occurs when a = 1/3, b = 2/3, c = 1, d = 4/3, e = 5/3, f = 2.

Using Lagrange's multipliers, one obtains $\large a = \frac{b}{2} = \frac{c}{3} = \frac{d}{4}= \frac{e}{5} = \frac{f}{6}$ .

Note that the denominator is nothing but the square root of the numerator of the respective terms.

This gives the value of a as $\frac{1}{3}$ . Substitute and simplify.

One could use AM-HM on the numbers used by @Aditya Chauhan