Ta có: \(a^3+b^3+ab=\left(a+b\right)^3-3ab\left(a+b\right)+ab\)

\(=1-3ab+ab=1-2ab=1-2\left(1-b\right)b\)

\(=1-2b+2b^2=2\left(b^2-b+\frac{1}{4}\right)+\frac{1}{2}\)

\(=2\left(b-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Đẳng thức xảy ra khi a = b \(=\frac{1}{2}\)

\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)

\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)

\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)

\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)

từ giả thiết, ta có \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

ta có \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\left(vi:\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\right)\) (ĐPCM)

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