Bài 3
2)
a). a^3 + b^3 >= a^2b + 2b^2a
<=> a^3 + b^3 - a^2b - 2b^2a >= 0
<=> a^2 ( a-b) - b^2( a-b) >= 0
<=> (a-b)^2 . (a+b) >= 0 => luôn đúng
b) \frac{1}{a^{3}+b^{3}+abc} + \frac{1}{b^{3}+c^{3}+abc} + \frac{1}{c^{3}+b^{3}+abc}\leq \frac{1}{abc}
Ta có:\frac{1}{a^{3}+b^{3}+abc} \leq...