MHB Inequality of Four Variables: Prove Σab(a^2+b^2+c^2)≤2

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The discussion centers on proving the inequality Σab(a^2+b^2+c^2)≤2 for non-negative real numbers a, b, c, and d constrained by the condition a + b + c + d = 2. Participants are encouraged to share their solutions or approaches to the problem. The inequality involves terms that combine products of the variables with their squares, indicating a relationship between their magnitudes. The conversation highlights the challenge of the proof and invites mathematical reasoning and techniques. Engaging with this inequality can enhance understanding of algebraic manipulation and inequalities in mathematics.
anemone
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Let $a,\,b,\,c$ and $d$ be non-negative real numbers such that $a + b + c + d = 2$.

Prove that $ab(a^2+ b^2 + c^2) + bc(b^2+ c^2+ d^2) + cd(c^2+ d^2+ a^2) + da(d^2+ a^2+ b^2) ≤ 2$.
 
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anemone said:
Let $a,\,b,\,c$ and $d$ be non-negative real numbers such that $a + b + c + d = 2$.

Prove that $ab(a^2+ b^2 + c^2) + bc(b^2+ c^2+ d^2) + cd(c^2+ d^2+ a^2) + da(d^2+ a^2+ b^2) ≤ 2$.

Please post the solution you have ready. (Time) (Wasntme)

You'd think I would've let you slide on this given that you've posted so many problems, but that's just not how I roll. (Bandit)
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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