For a,b,c>0, prove that a3a3+b3+abc+b3b3+c3+abc+c3c3+a3+abc≥1
I tried the following ∑cyca3a3+b3+abc⋅∑cyca3+b3+abc≥(a3/2+b3/2+c3/2)2
where cyc stands for cyclic sum of a,b,c, then we could show that (a3/2+b3/2+c3/2)2≥2(a3+b3+c3)+3abc to prove our original inequality. This looks similar to Schur but with inequality sign reversed. How would you proceed? Any ideas/hint?
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