Question

@mukushla

Prove \(ab + bc + ac \ge \sqrt{3abc \left(a+b+c\right)}\).

Answer 1

could u plz typr it again?

Answer 2

i am sorry, i re-did it. refresh.

Answer 3

ok

Answer 4

it just needed squaring and then it becomes,
\[a^2b^2 + c^2b^2 + a^2c^2 \ge abc(a+b+c)\]
and then a simple complete square. i act so foolishly sometimes, sorry for troubling you mukushla

Answer 5

squaring both sides
\[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)≥ 3abc(a+b+c)\\ a^2b^2+b^2c^2+c^2a^2-abc(a+b+c)\ge 0 \\2a^2b^2+2b^2c^2+2c^2a^2-2abc(a+b+c)\ge 0\\a^2b^2-2a^2bc+a^2c^2+a^2b^2-2ab^2c+b^2c^2+a^2c^2-2abc^2+b^2c^2\ge0\\ (ab−bc)^2+(bc−ca)^2+(ca−ab)^2\ge0\]