TITLE: Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$
QUESTION [5 upvotes]: Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove  $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$

The inequality can be written in the condensed form
$$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$
I was told that this is a pretty inequality to prove, but I have been unable to do so. 
I've tried naive things like multiplying both sides by $a+b+c+d$, and rewriting $(a^2+b^2+c^2+d^2)^2$, but nothing panned out (and the computations were relatively time-consuming). I also tried looking for clever applications of Cauchy-Schwarz (which seems like the way to go) and AM-GM, but nothing sprung out at me.

REPLY [2 votes]: instead of solving this problem let's solve a simpler version of it : 
Given a,b>0 and a^2+b^2=1, prove
a+b≥a^3+b^3+ab
you need to use what you have been given : 
we know a^2 + b^2 = 1 
adding 2ab to both sides of equation we have : 
a^2 +b^2 + 2ab = 1 + 2ab 
then : 
(a+b)^2 = 1 + 2ab (1) 
what we need to prove is that 
a+b≥a^3+b^3+ab (2)
by adding 3a^2b + 3ab^2 to the both sides of (2) we have : 
a + b + 3a^2b + 3ab^2 ≥ a^3 + b^3 + 3a^2b + 3ab^2 + ab (2) 
(3ab+1)(a+b) ≥ (a+b)^3 +ab 
using (1) we have 
(3ab+1)(a+b) ≥ (2ab+1)(a+b) +ab 
(ab)(a+b) ≥ ab
a+b ≥ 0 
true statement based on assumption 
so  
a+b≥a^3+b^3+ab
we can use the same procedure to solve the original problem in the case that : 
a^2+b^2+c^2+d^2=1   (1)
and then try to turn it to an equation which is simpler and easier to use in 
equation below 
a+b+c+d≥a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd  (2)
so by adding 2ab +2ac + 2cd + 2ad + 2bc + 2bd we turn (1) into 
a^2+b^2+c^2+d^2 + 2ab +2ac + 2cd + 2ad + 2bc + 2bd = 1 + 2ab +2ac + 2cd + 2ad + 
2bc + 2bd 
(a+b+c+d)^2 = 1 + 2ab +2ac + 2cd + 2ad + 2bc + 2bd (3)
then we use equation (3) to simplify equation (2) 
a+b+c+d
+ ab^2 + ac^2 + a*d^2 + 2a^2b +2a^2c + 2acd + 2a^2d + 2abc + 
ba^2+bc^2+bd^2 + 2ab^2 +2abc + 2bcd + 2abd + 2b^2c + 2b^2d+
ca^2+cb^2+cd^2 + 2abc +2ac^2 + 2c^2d + 2acd + 2bc^2 + 2bcd+
da^2+db^2+dc^2+ 2abd +2acd + 2cd^2 + 2ad^2 + 2bcd + 2bd^2
≥ 
a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd +
ab^2 + ac^2 + a*d^2 + 2a^2b +2a^2c + 2acd + 2a^2d + 2abc + 
ba^2+bc^2+bd^2 + 2ab^2 +2abc + 2bcd + 2abd + 2b^2c + 2b^2d+
ca^2+cb^2+cd^2 + 2abc +2ac^2 + 2c^2d + 2acd + 2bc^2 + 2bcd+
da^2+db^2+dc^2+ 2abd +2acd + 2cd^2 + 2ad^2 + 2bcd + 2bd^2 = 
(a+b+c+d)^3 +ab+ac+ad+bc+bd+cd = 
(1 + 2ab +2ac + 2cd + 2ad + 2bc + 2bd ) ( a+b+c+d) +ab+ac+ad+bc+bd+cd =
a + 2a^2b + 2a^2c + 2acd + 2a^2d + 2abc + 2abd +
b + 2ab^2 +2abc + 2bcd + 2abd + 2b^2c + 2b^2d+ 
c + 2abc +2ac^2 + 2c^2d + 2acd + 2bc^2 + 2bcd+ 
d + 2abd +2acd + 2cd^2 + 2ad^2 + 2bcd + 2bd^2+
+ab+ac+ad+bc+bd+cd
a+b+c+d 

ab^2 + ac^2 + a*d^2 + 2a^2b +2a^2c + 2acd + 2a^2d + 2abc + 

ba^2+bc^2+bd^2 + 2ab^2 +2abc + 2bcd + 2abd + 2b^2c + 2b^2d+
ca^2+cb^2+cd^2 + 2abc +2ac^2 + 2c^2d + 2acd + 2bc^2 + 2bcd+
da^2+db^2+dc^2+ 2abd +2acd + 2cd^2 + 2ad^2 + 2bcd + 2bd^2≥
a + 2a^2b + 2a^2c + 2acd + 2a^2d + 2abc + 2abd +
b + 2ab^2 +2abc + 2bcd + 2abd + 2b^2c + 2b^2d+ 
c + 2abc +2ac^2 + 2c^2d + 2acd + 2bc^2 + 2bcd+ 
d + 2abd +2acd + 2cd^2 + 2ad^2 + 2bcd + 2bd^2+
+ab+ac+ad+bc+bd+cd
after simplifying we reach to a true statement