Question

[Tường Nguyễn Thế]

Cho a, b, c à số dương thỏa mãn: ab+bc+ca=1. Tìm \(P_{min}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\sqrt{\dfrac{1}{a^2}+1}-\sqrt{\dfrac{1}{b^2}+1}-\sqrt{\dfrac{1}{c^2}+1}\)

Answer 1

[Lightning Farron]

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\end{matrix}\right.\)\(\Rightarrow x+y+z=xyz\)

\(\Rightarrow P=xy+yz+xz-\sqrt{x^2+1}-\sqrt{y^2+1}-\sqrt{z^2+1}\)

Khi \(a=b=c=\frac{1}{\sqrt{3}}\Rightarrow x=y=z=\sqrt{3}\Rightarrow P=3\)

Ta sẽ chứng minh \(P=3\) là giá tri nhỏ nhất của \(P\)

\(\Rightarrow BDT\Leftrightarrow xy+yz+xz-3\ge\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\)

Ta có BĐT \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2\)

\(\Leftrightarrow\left(xy+yz+xz\right)^2\ge x^2y^2z^2+2xyz\left(x+y+z\right)\)\(=3\left(x+y+z\right)^2\)

Xét \(VT^2=\left(xy+yz+xz-3\right)^2=\left(xy+yz+xz\right)^2-6\left(xy+yz+xz\right)+9\)

\(\ge3\left(x+y+z\right)^2-6\left(xy+yz+xz\right)+9\)\(=3\left(x^2+y^2+z^2\right)+9\left(1\right)\)

Và \(VP^2\le\left(1+1+1\right)\left(x^2+y^2+z^2+3\right)=3\left(x^2+y^2+z^2\right)+9\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) ta có ĐPCM. Vậy \(P_{min}=3\Rightarrow a=b=c=\frac{1}{\sqrt{3}}\)