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

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The inequality Σab(a^2+b^2+c^2)≤2 holds for non-negative real numbers a, b, c, and d constrained by the condition a + b + c + d = 2. The expression involves terms ab(a^2 + b^2 + c^2), bc(b^2 + c^2 + d^2), cd(c^2 + d^2 + a^2), and da(d^2 + a^2 + b^2). A rigorous proof is required to establish the validity of this inequality under the given constraints.

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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)
 

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