Lời giải:
Áp dụng BĐT Cô-si:
a^3+2b^3=a^3+b^3+b^3\geq 3\sqrt[3]{a^3b^6}=3ab^2$

$a^3+1+1\geq 3a$

$b^3+1+1\geq 3b$

Cộng theo vế các BĐT trên:

$a^3+2b^3+(a^3+2)+2(b^3+2)\geq 3ab^2+3a+6b$

$\Leftrightarrow 2(a^3+2b^3)+6\geq 3(ab^2+a+2b)=3.4=12$

$\Rightarrow a^3+2b^3\geq (12-6):2=3$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=1$

Ta có: (a-b)²+(b-c)²+(c-a)²=(a+b-2c)²+(b+c-2a)²+(c+a-2b)²=(a-c+b-c)²+(b-a+c-a)² +(c-b+a-b)².

Đặt a-b=x; b-c=y; c-a=z thì ta có:x+y+z=0,→ x²+y²+z²=(y-z)²+(z-x)²+(x-y)²=2(x² +y² +z²)-2(yz+xz+yx)

→x² +y² +z²+2(xy+yz+xz)=2(x² +y² +z²)

hay(x+y+z)²=2(x² +y² +z²). Mà x+y+z=0 nên→ x²+y²+z²=0,

→(a-b)²+(b-c)² +(c-a)²=0↔a-b=b-c=c-a=0→a=b=c(đpcm)

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z