Question

cho a+b≥0 chứng minh\(\frac{a+b}{2}\ge\sqrt[3]{\frac{a^3+b^3}{2}}\)

Answer 1

Lời giải:

Sửa lại đề. Cho $a+b\geq 0$. CMR \(\frac{a+b}{2}\leq \sqrt[3]{\frac{a^3+b^3}{2}}\)

Ta có:
\(a^3+b^3=(a+b)(a^2-ab+b^2)(1)\)

\(a^2-ab+b^2=(a+b)^2-3ab\)

\((a-b)^2\geq 0\Rightarrow a^2+b^2\geq 2ab\Rightarrow (a+b)^2\geq 4ab\Rightarrow \frac{3}{4}(a+b)^2\geq 3ab\)

\(\Rightarrow a^2-ab+b^2=(a+b)^2-3ab\geq (a+b)^2-\frac{3}{4}(a+b)^2=\frac{(a+b)^2}{4}(2)\)

Từ \((1);(2)\Rightarrow a^3+b^3\geq (a+b).\frac{(a+b)^2}{4}\)

\(\Rightarrow \frac{a^3+b^3}{2}\geq \frac{(a+b)^3}{8}\Rightarrow \sqrt[3]{\frac{a^3+b^3}{2}}\geq \frac{a+b}{2}\) (đpcm)

Dấu "=" xảy ra khi $a=b\geq 0$