MHB Problem #422: Four Non-Negative Real Numbers and a Unique Condition

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The problem involves four non-negative real numbers, a, b, c, and d, which satisfy the equation 2(ab+ac+ad+bc+bd+cd)+abc+abd+acd+bcd=16. The goal is to prove that a+b+c+d is at least 2/3 of the sum of the products ab, ac, ad, bc, bd, and cd. Additionally, the discussion seeks to identify the conditions under which equality holds. The lack of responses to the previous problem of the week highlights a potential engagement issue within the forum. The thread encourages participants to explore the mathematical relationships and proofs involved.
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Here is this week's POTW:

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Let $a,\,b,\,c$ and $d$ be four non-negative real numbers satisfying the condition

$2(ab+ac+ad+bc+bd+cd)+abc+abd+acd+bcd=16$

Prove that

$a+b+c+d\ge \dfrac{2}{3}(ab+ac+ad+bc+bd+cd)$

and determine when equality occurs.

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No one answered last week's POTW. (Sadface)

You can find the suggested solution as follows:
For $i= 1,\,2,\,3$, define $s_i$ as the average of the products of the i-element subsets of ${a,b,c,d}$. Then we must show

$3s_2+s_3=4\implies s_1\ge s_2$.

It suffices to prove the (unconstrained) homogeneous inequality

$3s_2^2s_1^2+s_3s_1^3\ge 4s_2^3$,

as then $3s_2+2_3=4$ will imply $(s_1-s_2)^3+3(s_1^3-s_2^3)\ge 0$.

We now recall two basic inequalities about symmetric means of non-negative real numbers. The first is Schur's inequality:

$3s_1^3+s_3\ge 4s_1s_2$,

while the second,

$s_1^2\ge s_2$ is a case of Maclaurin's inequality $s_i^{i+1}\ge s_{i+1}^i$.

These combine to prove the claim:

$3s_2^2s_1^2+s_3s_1^3\ge 3s_2^2s_1^2+\dfrac{s_2^2s_3}{s_1}\ge 4s_2^3$

Finally, for those who have only seen Schur's inequality in three variables, note that in general any inequality involving $s_1,\cdot,s_k$ which holds for $n\ge k$ variables also holds for $n+1$ variables, by replacing the variables $x_1,\cdots,x_{n+1}$ by the roots of the derivative of the polynomial $(x-x_1)\cdots (x-x_{n+1})$.
 
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