Question

[Phạm Băng Băng]

1. Cho \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) , trong đó: \(x,y,z>0\)

Chm: \(x=y=z\)

2. Cho \(a_1,a_2,...,a_n>0\) và \(a_1a_2...a_n=1\) Chm: \(\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right)\ge2^n\)

3. Chm \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge2\sqrt{2\left(a+b\right)\sqrt{ab}}\) \(\left(a,b\ge0\right)\)

Answer 1

[Phạm Minh Quang]

1. Ta có: \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Rightarrow\left(x+y+z\right)^2=\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=xy+yz+zx+2y\sqrt{xz}+2z\sqrt{xy}+2x\sqrt{yz}\)

\(\Leftrightarrow x^2+y^2+z^2+xy+yz+zx-2y\sqrt{xz}-2z\sqrt{xy}-2x\sqrt{yz}=0\)

\(\Leftrightarrow\left(x-\sqrt{yz}\right)^2+\left(y-\sqrt{xz}\right)^2+\left(z-\sqrt{xy}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{yz}\\y=\sqrt{xz}\\z=\sqrt{xy}\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\Rightarrow x=y=z\)

Answer 2

[Akai Haruma]

Bài 1:
\(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)

\(\Leftrightarrow x+y+z-\sqrt{xy}-\sqrt{yz}-\sqrt{xz}=0\)

\(\Leftrightarrow 2x+2y+2z-2\sqrt{xy}-2\sqrt{yz}-2\sqrt{xz}=0\)

\(\Leftrightarrow (x+y-2\sqrt{xy})+(y+z-2\sqrt{yz})+(z+x-2\sqrt{xz})=0\)

\(\Leftrightarrow (\sqrt{x}-\sqrt{y})^2+(\sqrt{y}-\sqrt{z})^2+(\sqrt{z}-\sqrt{x})^2=0\)

Vì \( (\sqrt{x}-\sqrt{y})^2;(\sqrt{y}-\sqrt{z})^2;(\sqrt{z}-\sqrt{x})^2\geq 0, \forall x,y,z>0\) nên để tổng của chúng bằng $0$ thì:

\( (\sqrt{x}-\sqrt{y})^2=(\sqrt{y}-\sqrt{z})^2=(\sqrt{z}-\sqrt{x})^2=0\)

\(\Rightarrow x=y=z\) (đpcm)