Please expand: (distrbute & multiply & add or subtract)a,b,c >0, proof that a^3+b^3+c^3+3abc ≥ a^2b+a^2c+b^2a+b^2c+c^2a+c^2b
Did you copy the question correctly?a,b,c >0, proof that a^3+b^3+c^3 + 3abc ≥ a^2b+a^2c+b^2a+b^2c+c^2a+c^2b
I was hoping (against hope) that OP will check that after expanding!!Did you copy the question correctly?
Subhotosh Khan's hint will apply if the sign in front of 3abc is negative.
My apologies. I failed to use mind-reading.I was hoping (against hope) that OP will check that after expanding!!
If we made the sign in front of 3abc negative then the inequality wouldn't be true.Did you copy the question correctly?
Subhotosh Khan's hint will apply if the sign in front of 3abc is negative.
Indeed. That's why I questioned whether SK's hint was applicable.If we made the sign in front of 3abc negative then the inequality wouldn't be true.
This problem was solved many many full-moons ago, in this forum. Since the OP did not show any work, I was trying nudge OP to try factorization. Beyond that, the problem statement was ill-formatted and that needed to be fixed.Indeed. That's why I questioned whether SK's hint was applicable.
I managed to prove the inequality using a few facts starting with expanding this one.
\(\displaystyle (a+b+c)(a^2+b^2+c^2)\)