Qui đồng chứng minh tương đương là ra
\(a+b=2c\Rightarrow\left\{{}\begin{matrix}c=\frac{a+b}{2}\\a-c=c-b\end{matrix}\right.\)
\(\frac{1}{\sqrt{a}+\sqrt{c}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{\sqrt{a}-\sqrt{c}}{a-c}+\frac{\sqrt{b}-\sqrt{c}}{b-c}=\frac{\sqrt{a}-\sqrt{c}}{a-c}-\frac{\sqrt{b}-\sqrt{c}}{a-c}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{a-c}=\frac{\sqrt{a}-\sqrt{b}}{a-\frac{a+b}{2}}=\frac{2\left(\sqrt{a}-\sqrt{b}\right)}{a-b}=\frac{2}{\sqrt{a}+\sqrt{b}}\)
Cách khác.
Đặt \(x=\frac{1}{\sqrt{a}+\sqrt{c}};y=\frac{1}{\sqrt{b}+\sqrt{c}};z=\frac{1}{\sqrt{a}+\sqrt{b}}\)(*)
Cần chứng minh \(x+y=2z\)
(*)\(\Leftrightarrow\frac{1}{x}=\sqrt{a}+\sqrt{c};\frac{1}{y}=\sqrt{b}+\sqrt{c};\frac{1}{z}=\sqrt{a}+\sqrt{b}\)
Cộng vế :
\(2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{x}+\sqrt{a}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow a=\frac{1}{4}\cdot\left(\frac{1}{y}+\frac{1}{z}-\frac{1}{x}\right)^2\)
Tương tự :
\(b=\frac{1}{4}\cdot\left(\frac{1}{x}-\frac{1}{y}+\frac{1}{z}\right)^2\)
\(c=\frac{1}{4}\cdot\left(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}\right)^2\)
Theo giả thiết : \(a+b=2c\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{x}-\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{4}\cdot\left[\left(\frac{1}{y}+\frac{1}{z}-\frac{1}{x}\right)^2+\left(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}\right)^2\right]\)
\(\Leftrightarrow\frac{4}{xy}-\frac{2}{yz}-\frac{2}{zx}=0\)
\(\Leftrightarrow\frac{2}{xy}=\frac{1}{yz}+\frac{1}{zx}\)
\(\Leftrightarrow\frac{2z}{xyz}=\frac{x+y}{xyz}\)
\(\Leftrightarrow2z=x+y\) ( đpcm )
@Trần Thanh Phương @Nguyễn Việt Lâm
