Prove that for any positive reals x, y, z we have (2x+y+z)2(2x2 + (y+z)2) + (2y+z+x)2(2y2 + (z+x)2) + (2z+x+y)2(2z2 + (x+y)2) ≤ 8.
Solution
If the inequality holds for x, y, z, then it also holds for kx, ky, kz, so it is sufficient to prove the result for x+y+z=3. The first term becomes (x+3)2/(2x2+(3-x)2) = (1/3) (x2+6x+9)/(x2-2x+3) = (1/3) (1 + (8x+6)/(2+(x-1)2) ≤ (1/3) (1 + (8x+6)/2) = 4/3 + 4x/3. Similarly for the other terms, so the whole expression ≤ (4/3 + 4x/3) + (4/3 + 4y/3) + (4/3 + 4z/3) = 8.
© John Scholes
jscholes@kalva.demon.co.uk
6 Aug 2003
Last corrected/updated 6 Aug 03