Có: A=\(\dfrac{1}{x}+\dfrac{1}{y}\) =\(\dfrac{x+y}{xy}\) =\(\dfrac{1}{xy}\) ( do x+y=1)
Áp dụng bđt \(\dfrac{a+b}{2}\ge\sqrt{ab}\) ,dâú bằng xảy ra khi a=b, ta có:
A=\(\dfrac{1}{x}+\dfrac{1}{y}\) =\(\dfrac{1}{xy}\) ≥ \(\dfrac{2}{x+y}\) =\(\dfrac{2}{1}\) =2 ( x+y=1)
dấu bằng xảy ra khi x=y=0,5.
c/m bđt \(\dfrac{a+b}{2}\ge\sqrt{ab}\) ⇔ a+b ≥ 2\(\sqrt{ab}\)
⇔(a+b)2 ≥ 4ab
⇔a2 +b2 +2ab≥ 4ab
⇔(a-b)2 ≥ 0 (luôn đúng)
dấu bằng xảy ra khi a=b.
\(\dfrac{a+b}{2}\ge\sqrt{ab}\left(\circledast\right)\\ \Leftrightarrow a+b\ge2\sqrt{ab}\\ \Leftrightarrow\left(a+b\right)^2\ge4ab\\ \Leftrightarrow a^2+2ab+b^2-4ab\ge0\\ \Leftrightarrow a^2-2ab+b^2=\left(a-b\right)^2\ge0\left(\text{luôn đúng}\right)\)
Vậy BĐT (*) được chứng minh.
\(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{1}{xy}\)
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\(\dfrac{x+y}{2}\ge\sqrt{xy}\\ \Rightarrow\sqrt{xy}\le\dfrac{1}{2}\\ \Rightarrow xy\le\dfrac{1}{4}\\ \Rightarrow A=\dfrac{1}{xy}\ge\dfrac{1}{\dfrac{1}{4}}=4\)
Vậy GTNN của A = 4
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
\(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\)
Theo đề bài, ta có:
\(\dfrac{x+y}{2}\ge\sqrt{xy}\\ \Leftrightarrow\dfrac{\left(x+y\right)^2}{4}\ge xy\\ \Leftrightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1^2}{4}=\dfrac{1}{4}\\ \Leftrightarrow xy\le\dfrac{1}{4}\\ \Leftrightarrow A\le\dfrac{x+y}{xy}=\dfrac{1}{\dfrac{1}{4}}=4\)
Vậy \(A_{min}=4\Leftrightarrow x=y=\dfrac{1}{2}\)
Ta có: \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge0\) \(\forall a,b\ge0\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\) \(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\) (đpcm) \(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(A=\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}\)
Mặt khác: \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)
\(\Rightarrow A\ge2\sqrt{4}=4\)
Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy \(Min_A=4\) khi \(x=y=\dfrac{1}{2}\)
