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

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.

Answer 1

Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $abc+abd+acd+bcd=4w^3$.

Hence, $16u^2-12v^2=1$ and our inequality is equivalent to $3v^6-4uv^2w^3+w^6\geq0$.

By Roll's theorem there are $x>0$, $y>0$ and $z>0$, for which

$x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

After this substitution we need to prove that

$\sum\limits_{cyc}(x^3y^3-x^3y^2z-x^3z^2y+x^2y^2z^2)\geq0$, which is Schur.

Answer 2

I got some hints from Crux Problem 3059, click here for more details=)

Based on the hints (which tries to relate the inequality with a constrained optimization problem), I worked out a proof as follows:

We have \begin{align} \frac{1}{2}(a+b+c+d)^2=&\frac{1}{2}(a^2+b^2+c^2+d^2)+ab+ac+ad+bc+bd+cd\\ =&\frac{1}{2}+ab+ac+ad+bc+bd+cd \end{align} by assumption. Thus, in order to prove the inequality, it is equivalent to prove $$ a+b+c+d\geq a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d)^2-\frac{1}{2}. $$ This can further be simplified as proving $$ a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d-1)^2\leq1.~~~~(*) $$ Now we try to maximize the LHS under the constraint, i.e. \begin{align} \max&~~f\triangleq a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d-1)^2\\ s.t.&~~a^2+b^2+c^2+d^2=1. \end{align} Now we try to use the Lagrangian multiplier method. Let $$ L=f+\lambda(a^2+b^2+c^2+d^2-1). $$ Take the derivative of $L$ regarding $a,b,c,d$ respectively and let the derivative equal to $0$ gives the following set of equations: \begin{align} L_a=&a+b+c+d+3a^2+2a\lambda=1\\ L_b=&a+b+c+d+3b^2+2b\lambda=1\\ L_c=&a+b+c+d+3c^2+2c\lambda=1\\ L_d=&a+b+c+d+3d^2+2d\lambda=1, \end{align}

Notice that these four equations share exactly the same form. Thus, either $$ 3x^2+2\lambda x=0,~~x=a,b,c,d,~~~~(case1) $$ or $$ a=b=c=d.~~~~(case2) $$ We analyse these two cases separately:

Case 1: Under this case, we have $$ a+b+c+d=1. $$ Since $$ a^2+b^2+c^2+d^2=1, $$ the only possibility is one of four elements $a,b,c,d$ equal to $1$ and others are all $0$, which gives $\lambda=-3/2$ and $f=1$.

Case 2: By the assumption $$ a^2+b^2+c^2+d^2=1, $$ it follows $$ a=b=c=d=\frac{1}{2}, $$ which gives $\lambda=-7/4$ and $f=1$.

Above all we have the maximum of $f$ is $1$ and this proves $(*)$.

Answer 3

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