Question

(a^3+b^3)/2 >= ((a+b)/2)^2

Answer 1

Sửa đề: \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)

=>\(\frac{a^3+b^3}{2}\ge\frac{\left(a+b\right)^3}{8}\)

=>\(4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)

=>\(4\left(a+b\right)\left(a^2-ab+b^2\right)\ge\left(a+b\right)^3\)

=>\(\left(a+b\right)\cdot\left(4a^2-4ab+4b^2-a^2-2ab-b^2\right)\ge0\)

=>\(\left(a+b\right)\left(3a^2-6ab+3b^2\right)\ge0\)

=>\(3\left(a+b\right)\left(a^2-2ab+b^2\right)\ge0\)

=>\(\left(a+b\right)\left(a-b\right)^2\ge0\) (luôn đúng với a,b dương)

=>ĐPCM