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Ta có: \(\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng)
Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng )
Dấu "=" xảy ra khi \(a=b\)
Bài tập :
Có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{x}+\dfrac{x+y}{y}=2+\dfrac{x}{y}+\dfrac{y}{x}\) ( do \(x+y=1\) )
Theo BĐT trên có : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2.\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)
Nên \(A\ge2+2=4\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
*Chứng minh bất đẳng thức
Ta có: \(\forall a,b\ge0\) thì \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\) \(\Leftrightarrow a+b\ge2\sqrt{ab}\) \(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\) (đpcm)
Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\forall a,b>0\)
\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\forall a,b>0\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\forall a,b>0\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\forall a,b>0\)(đpcm)
Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\) (tự cm)
Lại có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\)
Áp dụng BĐT trên ta có : : \(xy\le\left(\dfrac{x+y}{2}\right)^2\)
\(\Leftrightarrow A\ge\dfrac{x+y}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{1}{\dfrac{1}{2^2}}=4\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy...
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 \(\Leftrightarrow x=y=\dfrac{1}{2}\)
\(P=\left(a^2+\dfrac{1}{16a^2}\right)+\left(b^2+\dfrac{1}{16b^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge2\sqrt{\dfrac{a^2}{16a^2}}+2\sqrt{\dfrac{b^2}{16b^2}}+\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)
\(P\ge1+\dfrac{15}{32}.\left(\dfrac{4}{a+b}\right)^2\ge1+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=\dfrac{17}{2}\)
\(P_{min}=\dfrac{17}{2}\) khi \(a=b=\dfrac{1}{2}\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
Đặt \(\left(2\sqrt{a}-5;2\sqrt{b}-5;2\sqrt{c}-5\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x;y;z>0\\a=\left(\dfrac{x+5}{2}\right)^2\\b=\left(\dfrac{y+5}{2}\right)^2\\c=\left(\dfrac{z+5}{2}\right)^2\end{matrix}\right.\)
\(Q=\dfrac{\left(x+5\right)^2}{4y}+\dfrac{\left(y+5\right)^2}{4z}+\dfrac{\left(z+5\right)^2}{4x}\ge\dfrac{\left(x+y+z+15\right)^2}{4\left(x+y+z\right)}\)
\(Q\ge\dfrac{\left(x+y+z\right)^2+30\left(x+y+z\right)+225}{4\left(x+y+z\right)}\)
\(Q\ge\dfrac{x+y+z}{4}+\dfrac{225}{4\left(x+y+z\right)}+\dfrac{15}{2}\ge2\sqrt{\dfrac{225\left(x+y+z\right)}{16\left(x+y+z\right)}}+\dfrac{15}{2}=15\)
Dấu "=" xảy ra khi \(a=b=c=25\)
Áp dụng bđt hoán vị cho hai bộ số đơn điệu ngược chiều \(\left(a,b,c\right);\left(2\sqrt{a}-5,2\sqrt{b}-5,2\sqrt{c}-5\right)\): \(Q\ge\dfrac{a}{2\sqrt{a}-5}+\dfrac{b}{2\sqrt{b}-5}+\dfrac{c}{2\sqrt{c}-5}\).
Mặt khác ta có \(\dfrac{a}{2\sqrt{a}-5}-5=\dfrac{\left(\sqrt{a}-5\right)^2}{2\sqrt{a}-5}\ge0\).
Do đó \(Q\ge5+5+5=15\).
Dấu bằng xảy ra khi a = b = c = 25.
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)
\(\Rightarrow3.P\ge9\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Mình không biết làm :>
Áp dụng BĐT Cauchy Schwarz có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}\ge\dfrac{4}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}\left(vìa+b\le2\sqrt{2}\right)\)
Dấu ''='' xảy ra khi \(a=b=\sqrt{2}\)