Question

If $a,b,c$ denote positive quantities, prove that
$a^2+b^2+c^2>bc+ca+ab;$
and $2(a^2+b^2+c^2)>bc(b+c)+ca(c+a)+ab(a+b)$.

Answer 1

for any two real numbers,sum of their squares is greater than twice their product

hence, $b^2+c^2>2bc....\left(i\right)$
$c^2+a^2>2ca........\left(ii\right)$
$a^2+b^2>2ab.......\left(iii\right)$

adding $i,ii,andiii$

$a^2+b^2+c^2>bc+ca+ab$,hence proved
It may be noticed that this result is true for any real values of $a,b,c$.

now,subtracting bc on both side of (i)
$b^2-bc+c^2>bc$
$\therefore b^2+c^2>bc(b+c)....\left(iv\right)$

similarly, $c^2+a^2>ac(a+c).......\left(v\right)$

$a^2+b^2>ab(a+b).......\left(vi\right)$
by adding $\left(iv\right),\left(v\right),\left(vi\right)$
$2(a^2+b^2+c^2)>bc(b+c)+ca(c+a)+ab(a+b)$,

hence proved