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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:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
1) Đặt \(\dfrac{b\sqrt{a-1}+a\sqrt{b-1}}{ab}\) là A
\(\)\(A=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-1}}{b}\)
\(\left(\dfrac{\sqrt{a-1}}{a}\right)^2=\dfrac{a-1}{a^2}=\dfrac{1}{a}-\dfrac{1}{a^2}=\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)\)
\(\Rightarrow\)\(\dfrac{\sqrt{a-1}}{a}=\sqrt{\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)}\)
Tương tự: \(\dfrac{\sqrt{b-1}}{b}=\sqrt{\dfrac{1}{b}\left(\dfrac{1}{b}-1\right)}\)
Áp dụng BĐT Cauchy, ta có:
\(\sqrt{\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)}\le\dfrac{\dfrac{1}{a}+\left(1-\dfrac{1}{a}\right)}{2}=\dfrac{1}{2}\)
Tương tự: \(\sqrt{\dfrac{1}{b}\left(\dfrac{1}{b}-1\right)}\le\dfrac{1}{2}\)
Cộng vế theo vế của 2 BĐT vừa chứng minh, ta được:
\(A\le1\left(đpcm\right)\)
Xét: \(a^2+\dfrac{2}{a^3}=\dfrac{1}{3}a^2+\dfrac{1}{3}a^2+\dfrac{1}{3}a^2+\dfrac{1}{a^3}+\dfrac{1}{a^3}\left(1\right)\)
Áp dụng BĐT Cauchy cho 5 số dương trên, ta có: \(\left(1\right)\ge5\sqrt[5]{\dfrac{1}{3}a^2.\dfrac{1}{3}a^2.\dfrac{1}{3}a^2.\dfrac{1}{a^3}.\dfrac{1}{a^3}}=5\sqrt[5]{\dfrac{1}{27}}=\dfrac{5\sqrt[5]{9}}{3}\left(đpcm\right)\)
Dấu ''='' xảy ra khi và chỉ khi \(\dfrac{1}{3}a^2=\dfrac{1}{a^3}\Leftrightarrow a=\sqrt[5]{3}\)
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
a: \(a+\dfrac{1}{a}\ge2\sqrt{a\cdot\dfrac{1}{a}}=2\)
b: \(\Leftrightarrow\dfrac{a^2+a+1+1}{\sqrt{a^2+a+1}}>=2\)
=>\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\)(1)
\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\sqrt{\sqrt{a^2+a+1}\cdot\dfrac{1}{\sqrt{a^2+a+1}}}=2\)
nên (1) đúng
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)
\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)
b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)
c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)