toán khó cần gấp

M

mimibili

cho a,b dương, [TEX]A=\frac{a+b}{2}; B=sqrt{ab}[/TEX]
CMR: [TEX]sqrt{ab} \leq P= \frac{(a-b)^2}{8.(\frac{a+b}{2}-sqrt{ab})} [/TEX] \leq [tex]\frac{a+b}{2}[/tex]
Ta có: [TEX]P=\frac{(\sqrt{a}-\sqrt{b})^2(\sqrt{a}+\sqrt{b})^2}{4.(\sqrt{a}-\sqrt{b})^2}=\frac{\sqrt{a}+\sqrt{b})^2}{4}=\frac{a+b+2\sqrt{ab}}{4}[/TEX]
\Rightarrow[tex] P\geq \frac{2\sqrt{ab}+2\sqrt{ab}}{4}=B[/tex]
và [TEX]P\leq\frac{a+b+a+b}{4}=A[/TEX]
Dấu = xảy ra khi chỉ khi a=b
 
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