1−a+1−b+1−c≤2(ab+bc+ca+2a2+b2+c2). for non-negative a,b,c with a+b+c=1.
Solution
From the Cauchy-Schwartz inequality for the collections (α,β,γ) and (αx,βy,γz) we obtain: (αx+βy+γz)≤(α+β+γ)(αx+βy+γz), and so aa+b+bb+c+cc+a≤aa+b+bb+c+cc+a≤(a+b+c)(a2+ab+b2+bc+c2+ca)≤2(a2+b2+c2). similarly, ba+b+cb+c+ac+a≤2(a2+b2+c2),ca+b+ab+c+bc+a≤2(ab+bc+ca).
Adding all three inequalities, we get the required one.
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Source: MathNet,
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