Cho \(x,y,z>0\) và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\). Chứng minh:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\ge1\)
@Đời về cơ bản là buồn... cười!!!, giúp tuôi!
+ Áp dụng bđt : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a,b>0\)
Dấu "=" xảy ra \(\Leftrightarrow a=b\) ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) . Dấu "=" xảy ra \(\Leftrightarrow x=y\)
+ Tương tự : \(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{4}{y+z}\). Dấu "=" xảy ra \(\Leftrightarrow y=z\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{4}{x+z}\). Dấu "=" xảy ra \(\Leftrightarrow x=z\)
Do đó : \(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{x+z}\)
\(\Rightarrow8\ge4\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\)
\(\Rightarrow2\ge\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
+ \(\dfrac{1}{x+y}+\dfrac{1}{y+z}\ge\dfrac{4}{x+2y+z}\). Dấu "=" xảy ra \(\Leftrightarrow x=z\)
\(\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{4}{x+y+2z}\). Dấu "=" xảy ra \(\Leftrightarrow x=y\)
\(\dfrac{1}{x+y}+\dfrac{1}{x+z}\ge\dfrac{4}{2x+y+z}\). Dấu "=" xảy ra \(\Leftrightarrow y=z\)
Do đó : \(2\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\ge\dfrac{4}{2x+y+z}\)
\(+\dfrac{4}{x+2y+z}+\dfrac{4}{x+y+2z}\)
\(\Rightarrow4\ge4\left(\dfrac{1}{x+2y+z}+\dfrac{1}{2x+y+z}+\dfrac{1}{x+y+2z}\right)\)
\(\Rightarrow1\ge\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=\dfrac{4}{3}\)
\(\Leftrightarrow x=y=z=\dfrac{4}{3}\)
* CM bđt : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a,b>0\)
+ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
Vì bđt cuối luôn đúng mà các phép biển đổi trên là tương đương nên bđt đầu luôn đúng
Dấu "=" xảy ra <=> a= b
