Ta có: $x^5+(1-x)^5-\frac{1}{16}$$=\frac{15}{16}-5x+10x^2-10x^3+5x^4$$=\frac{5}{16}(2x-1)^2(4x^2-4x+3)\ge0$Dấu bằng xảy ra khi: $x=\frac{1}{2}$
Ta có: $x^5+(1-x)^5-\frac{1}{16}$$=\frac{15}{16}-5x+10x^2-10x^3+5x^4$$=\frac{5}{16}(2x-1)^2(4x^2-4x+3)\le0$Dấu bằng xảy ra khi: $x=\frac{1}{2}$
Ta có: $x^5+(1-x)^5-\frac{1}{16}$$=\frac{15}{16}-5x+10x^2-10x^3+5x^4$$=\frac{5}{16}(2x-1)^2(4x^2-4x+3)\
ge0$Dấu bằng xảy ra khi: $x=\frac{1}{2}$