Prove that for all a,b,c>0 such that a+b+c≤3 the following inequality holds a(a+2)a+1+b(b+2)b+1+c(c+2)c+1≥2. (Tangenta magazine, 2010)
Solution — click to reveal
From the A–H inequality we get a(a+2)a+1+b(b+2)b+1+c(c+2)c+1≥a+1a(a+2)+b+1b(b+2)+c+1c(c+2)9=a+1(a+1)2−1+b+1(b+1)2−1+c+1(c+1)2−19=(a+1)+(b+1)+(c+1)−(a+11+b+11+c+11)9 Since (a+1)+(b+1)+(c+1)=a+b+c+3≤6 and a+11+b+11+c+11≥(a+1)+(b+1)+(c+1)9≥69=23 we can conclude (a+1)+(b+1)+(c+1)−(a+11+b+11+c+11)9≥6−239=2 which finishes the proof.