>>922
a, b, c, r > 0 に対して　(ab)^{r+1/2} (aa+bb-2cc) ≧ (ab-cc) (a+b) (cc)^r,

(略証)
（左辺) - (右辺)
≧ (a+b)(ab)^{r+1} - (a+b)(ab)^r･cc - (ab-cc)(a+b)(cc)^r
= (a+b)(ab-cc) [(ab)^r - (cc)^r]
≧ 0,