An algebra problem by Achal Jain

By AM - GM Inequality ,

$a^{2} + 1 \geq 2a$

$\Rightarrow \dfrac{1}{a^2 + 1} \leq \dfrac{1}{2a}$

$\Rightarrow \dfrac{a^{2}}{a^{2} + 1} \leq \dfrac{a}{2}$ .

Similarly writing for $\left(b , c , d\right)$ and adding all we get ,

$\dfrac{a^{2}}{a^{2} + 1} + \dfrac{b^{2}}{b^{2} + 1} + \dfrac{c^{2}}{c^{2} + 1} + \dfrac{d^{2}}{d^{2} + 1} \leq \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{c}{2} + \dfrac{d}{2} = 2$

Therefore ,

$\dfrac{-a^{2}}{a^{2} + 1} + \dfrac{-b^{2}}{b^{2} + 1} + \dfrac{-c^{2}}{c^{2} + 1} + \dfrac{-d^{2}}{d^{2} + 1} \geq \dfrac{-a}{2} + \dfrac{-b}{2} + \dfrac{-c}{2} + \dfrac{-d}{2} = -2$

$1 - \dfrac{a^{2}}{a^{2} + 1} + 1 - \dfrac{b^{2}}{b^{2} + 1} + 1 - \dfrac{c^{2}}{c^{2} + 1} + 1 - \dfrac{d^{2}}{d^{2} + 1} \geq 4 - 2 = 2$ .

$\Rightarrow \dfrac { 1 }{ { a }^{ 2 }+1 } +\dfrac { 1 }{ { b }^{ 2 }+1 } +\dfrac { 1 }{ { c }^{ 2 }+1 } + \dfrac { 1 }{ { d }^{ 2 }+1 } \geq 2$ .

Equality holds when $a = b = c = d = 1$

Relevant wiki: Titu's Lemma

This can be done by Titu's Lemma

Now here by applying Titu's Lemma we get,

$\frac{ 1 }{{ a }^{ 2 }+1 } +\frac { 1 }{ { b }^{ 2 }+1 } +\frac { 1 }{ { c }^{ 2 }+1 } + \frac { 1 }{ { d }^{ 2 }+1 }$ >= $\frac{4^2}{a^2+b^2+c^2+d^2+4}$

Now $a+b+c+d=4$ Hence by AM GM we get $abcd<=1$

And $a^2+b^2+c^2+d^2>=4$ .

Hence putting value in the 1st equation we get, $4^2/8= 2$