# How to show that $a^3+b^3+c^3+d^3\geq abc+abd+acd+bcd$ if $a,b,c,d>0$

How can I prove that if $a,b,c,d>0$ then

$$a^3+b^3+c^3+d^3\geq abc+abd+acd+bcd?$$

I think there is some simple proof but I can't remember... is this a special case of some general inequality?

I need it to prove that if $a^3+2c=3ab$ then all of 4 positive roots of following polynomial is the same.

$x^4+ax^3+bx^2+cx+d$

• Mathematical formulae look better in $\LaTeX$. Here is a quick tutorial. Apr 11, 2016 at 21:15
• Are you sure the LHS is correct? Apr 11, 2016 at 21:17
• I think LHS should be $a^3+b^3+c^3+d^3$...
– user327929
Apr 11, 2016 at 21:19
• If what's up there is correct, then the LHS simplifies down to $3a^3+b^3$, which means I could easily choose $c,d$ so that this inequality is false. Apr 11, 2016 at 21:21
• I assumed that BBB was correct about the LHS and edited accordingly; if that is wrong, you can fix it. See meta.math.stackexchange.com/questions/5020/… for explanations of the markup I used. Apr 11, 2016 at 21:26

hint: Use AM-GM inequality $4$ times:
$$x^3+y^3+z^3 \geq 3xyz$$
with $(x,y,z) = (a,b,c), (a,c,d), (a,b,d), (b,c,d)$, and add up.