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\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
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 \(a=b=c=\sqrt{3}\)
https://www.facebook.com/OnThiDaiHocKhoiA/posts/508217699295984
Áp dụng BĐT Bunhia:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{\left(1+1+1\right)\left(4a+1+4b+1+4c+1\right)}\)
\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3.\left(4\left(a+b+c\right)+3\right)}=\sqrt{21}< \sqrt{25}=5\)
Vậy \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)