Question

Show that $(a+b+c+d)(a^3+b^3+c^3+c^3)>{(a^2+b^2+c^2+c^2)}^2$.

Answer 1

Let assume $S=(a+b+c+d)(a^3+b^3+c^3+d^3)>{(a^2+b^2+c^2+d^2)}^2$

$S=(a+b+c+d)(a^3+b^3+c^3+d^3)-{(a^2+b^2+c^2+d^2)}^2>0$

$S=(a^4+b^4+c^4+d^4+a{b}^3+a{c}^3+a{d}^3+b{a}^3+b{c}^3+b{d}^3+c{a}^3+c{b}^3+c{d}^3+d{a}^3+d{b}^3+d{c}^3)$

$-(a^4+b^4+c^4+d^4+2(a^2{b}^2+a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2+c^2{d}^2))>0$

$S=(a{b}^3+b{a}^3-2a^2{b}^2)+(a{c}^3+c{a}^3-2a^2{c}^2)+(a{d}^3+d{a}^3-2a^2{d}^2)+(b{c}^3+c{b}^3-2c^2{b}^2)$

$+(b{d}^3+d{b}^3-2d^2{b}^2)+(c{d}^3+d{c}^3-2c^2{d}^2)>0$

$S=ab{(a-b)}^2+ac{(a-c)}^2+ad{(a-d)}^2+bc{(b-c)}^2+bd{(b-d)}^2+cd{(c-d)}^2>0$

As $a,b,c,d$ are positive thus square of every term is positive .

Therefore $S>0$