An algebra problem by swastik p (7)

Let's focus on $(a+b+c+d)\ge \text{?}$

By AM-GM inequality we have:

$\dfrac{a+b+c+d}{4}\ge \sqrt[4]{abcd}$

$\therefore a+b+c+d\ge 4\sqrt[4]{abcd} ~-~ \text{(i)}$

Now we focus on $\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \right)\ge \text{?}$

By AM-GM inequality we have:

$\dfrac{\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \right)}{4}\ge \sqrt[4]{ \dfrac{1}{a}\cdot \dfrac{1}{b}\cdot \dfrac{1}{c}\cdot \dfrac{1}{d}}$

$\therefore \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \ge 4\sqrt[4]{\dfrac{1}{abcd}}~-~ \text{(ii)}$

By $\text{(i)}\times \text{(ii)}$ we have

$\begin{aligned}\\& (a+b+c+d)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \right)\ge 4\sqrt[4]{abcd}\times 4\sqrt[4]{\dfrac{1}{abcd}} \\& \implies (a+b+c+d)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \right)\ge 16\sqrt[4]{\dfrac{abcd}{abcd}} \\& \implies \boxed{~~ (a+b+c+d)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \right)\ge 16 ~~}\end{aligned}$

Hence, the inequality is TRUE

The Cauchy-Schwarz inequality tells us that $({ x }_{ 1 }^{ 2 }+...+{ x }_{ n }^{ 2 })({ y }_{ 1 }^{ 2 }+...+{ y }_{ n }^{ 2 })\ge { ({ x }_{ 1 }{ y }_{ 1 }+...+{ x }_{ n }{ y }_{ n }) }^{ 2 }$ .

Plugging in ${ x }_{ 1 }=\sqrt { a }$ , ${ x }_{ 2 }=\sqrt { b }$ , ${ x }_{ 3 }=\sqrt { c }$ , ${ x }_{ n }={ x }_{ 4 }=\sqrt { d }$ , ${ y }_{ 1 }=\frac { 1 }{ \sqrt { a } }$ , ${ y }_{ 2 }=\frac { 1 }{ \sqrt { b } }$ , ${ y }_{ 3 }=\frac { 1 }{ \sqrt { c } }$ , and ${ y }_{ n }={ y }_{ 4 }=\frac { 1 }{ \sqrt { d } }$ , we get that $({ \sqrt { a } }^{ 2 }+{ \sqrt { b } }^{ 2 }+{ \sqrt { c } }^{ 2 }+{ \sqrt { d } }^{ 2 })(\frac { 1 }{ { \sqrt { a } }^{ 2 } } +\frac { 1 }{ { \sqrt { b } }^{ 2 } } +\frac { 1 }{ { \sqrt { c } }^{ 2 } } +\frac { 1 }{ { \sqrt { d } }^{ 2 } } )\ge { (\sqrt { a } \cdot \frac { 1 }{ \sqrt { a } } +\sqrt { b } \cdot \frac { 1 }{ \sqrt { b } } +\sqrt { c } \cdot \frac { 1 }{ \sqrt { c } } +\sqrt { d } \cdot \frac { 1 }{ \sqrt { d } } ) }^{ 2 }$ , which simplifies to $(a+b+c+d)(\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } )\ge 16$ .

This new equality is the same one used in the problem, which implies that, yes , the inequality holds true for all $a$ , $b$ , $c$ , and $d$ .