2015/088)If a,b,c are real and distinct, then show that

$a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\gt6abc$

taking 6abc to the left and reordering we get

$(a^2b^2-2bac + c^2) + (a^2c^2-2abc + b^2) + (b ^2c^2- 2abc + a^2)$

= $(ab-c)^2 + (ac-b)^2 + (bc-a)^2 \ge 0$

the expression can be zero only if $ab= c, ac = b , bc = a$ or $abc = $1 or $zero$

if a,b,c all are different then above cannot be zero