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Lời giải:
a) Ta thấy: \(a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0, \forall a,b>0\)
\(\Rightarrow a+b\geq 2\sqrt{ab}>0\Rightarrow \frac{1}{a+b}\le \frac{1}{2\sqrt{ab}}\).
Vì $a> b$ nên dấu bằng không xảy ra . Tức \(\frac{1}{a+b}< \frac{1}{2\sqrt{ab}}\)
Ta có đpcm
b)
Áp dụng kết quả phần a:
\(\frac{1}{3}=\frac{1}{1+2}< \frac{1}{2\sqrt{2.1}}\)
\(\frac{1}{5}=\frac{1}{3+2}< \frac{1}{2\sqrt{2.3}}\)
\(\frac{1}{7}=\frac{1}{4+3}< \frac{1}{2\sqrt{4.3}}\)
.....
\(\frac{1}{4021}=\frac{1}{2011+2010}< \frac{1}{2\sqrt{2011.2010}}\)
Do đó:
\(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{\sqrt{2}-\sqrt{1}}{2\sqrt{2.1}}+\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3.2}}+\frac{\sqrt{4}-\sqrt{3}}{2\sqrt{4.3}}+....+\frac{\sqrt{2011}-\sqrt{2010}}{2\sqrt{2011.2010}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2010}}-\frac{1}{2\sqrt{2011}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{2011}}< \frac{1}{2}\) (đpcm)
Câu a
\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right):\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{1}\)
\(=a-b\)
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
Cho a,b>0 thoã mãn: \(\dfrac{1}{a}+\dfrac{1}{b}=1\)
Chứng minh: \(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}=1\)
\(\dfrac{b}{ab}+\dfrac{a}{ab}=1\)
\(\dfrac{a+b}{ab}=1\)
\(\Rightarrow a+b=ab\)
Xét:
\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
\(\Leftrightarrow\left(\sqrt{a+b}\right)^2=\left(\sqrt{a-1}+\sqrt{b-1}\right)^2\)
\(\Leftrightarrow a+b=a-1+b-1+2\sqrt{\left(a-1\right)\left(b-1\right)}\)
\(\Leftrightarrow a+b=a+b-2+2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow2=2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow4=4\left(ab-a-b+1\right)\)
\(\Leftrightarrow4-4\left(ab-a-b+1\right)=0\)
\(\Leftrightarrow4\left(ab-a-b\right)=0\)
\(\Leftrightarrow ab-a-b=0\)
\(\Leftrightarrow ab=a+b\)
Mà bên trên đã cm đc ab=a+b nên đẳng thức trên luôn đúng
Hay :\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
=.= hok tốt !!
Ta có \(\dfrac{1}{a}+\dfrac{1}{b}=1\Leftrightarrow\dfrac{a+b}{ab}=1\Leftrightarrow a+b=ab\Leftrightarrow ab-a-b=0\Leftrightarrow1=ab-a-b+1\Leftrightarrow1=a\left(b-1\right)-\left(b+1\right)\Leftrightarrow1=\left(a-1\right)\left(b-1\right)\Leftrightarrow1=\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow0=-2+2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow a+b=a-1+2\sqrt{\left(a-1\right)}\sqrt{\left(b-1\right)}+b-1\Leftrightarrow a+b=\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\Leftrightarrow\sqrt{a+b}=\sqrt{\left(\sqrt{a+1}+\sqrt{b+1}\right)^2}\Leftrightarrow\sqrt{a+b}=\sqrt{a+1}+\sqrt{b+1}\left(đpcm\right)\)
\(\dfrac{1}{a}+\dfrac{1}{b}=1\)
\(\Leftrightarrow a+b=ab\)(*)
Xét
\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
\(\Leftrightarrow a+b=a+b-2+2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow2=2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow4=4\left(ab-a-b+1\right)\)
\(\Leftrightarrow4\left(ab-a-b\right)=0\Leftrightarrow ab-a-b=0\)
\(\Leftrightarrow ab=a+b\)(đúng với *)
\(\Rightarrow\)đpcm
Ta có\(\dfrac{1}{a}+\dfrac{1}{b}=1\Leftrightarrow\dfrac{a+b}{ab}=1\Leftrightarrow a+b=ab\Leftrightarrow ab-a-b=0\Leftrightarrow1=ab-a-b+1\Leftrightarrow1=a\left(b-1\right)-\left(b+1\right)\Leftrightarrow1=\left(a-1\right)\left(b-1\right)\Leftrightarrow1=\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow0=-2+2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow a+b=a-1+2\sqrt{\left(a-1\right)}\sqrt{\left(b-1\right)}+b-1\Leftrightarrow a+b=\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\Leftrightarrow\sqrt{a+b}=\sqrt{\left(\sqrt{a+1}+\sqrt{b+1}\right)^2}\Leftrightarrow\sqrt{a+b}=\sqrt{a+1}+\sqrt{b+1}\left(đpcm\right)\)