chứng minh bất dẳng thức sau
\(\frac{1}{a}\)\(+\frac{1}{b}\)bé hơn hoặc bằng \(\frac{2}{1+\sqrt{ab}}\)
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26 tháng 6 2016
3a) ta có \(\frac{a^2}{a+b}=a-\frac{ab}{a+b}>=a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)
vì \(a,b>0,a+b>=2\sqrt{ab}nên\frac{ab}{a+b}< =\frac{ab}{2\sqrt{ab}}\)
tương tự \(\frac{b^2}{b+c}=b-\frac{bc}{b+c}>=b-\frac{bc}{2\sqrt{bc}}=b-\frac{\sqrt{bc}}{2}\)
tương tự \(\frac{c^2}{c+a}=c-\frac{ca}{c+a}>=c-\frac{ca}{2\sqrt{ca}}=c-\frac{\sqrt{ca}}{2}\)
cộng từng vế BĐT ta được \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=a+b+c-\frac{\sqrt{ab}}{2}-\frac{\sqrt{bc}}{2}-\frac{\sqrt{ca}}{2}=\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}\left(1\right)\)
giả sử \(\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}>=\frac{a+b+c}{2}\)
<=> \(2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=a+b+c\)
<=> \(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=0\)
<=> \(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}>=0\)
<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)
(đúng với mọi a,b,c >0) (2)
(1),(2)=> \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=\frac{a+b+c}{2}\left(đpcm\right)\)
28 tháng 12 2015
Ta có
\(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\ge0\)
<=>\(a+b\ge2\sqrt{ab}\)
Dấu ''='' xảy ra <=>\(\sqrt{a}-\sqrt{b}=0<=>\sqrt{a}=\sqrt{b}<=>a=b\)
Tick cho tui nha,bạn hiền
28 tháng 12 2015
\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow a+b\ge2\sqrt{ab}\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}\)
TN
17 tháng 8 2017
a)Áp dụng BĐT AM-GM ta có
\(\frac{ab\sqrt{ab}}{a+b}\le\frac{ab\sqrt{ab}}{2\sqrt{ab}}=\frac{ab}{2}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\frac{bc\sqrt{bc}}{b+c}\le\frac{bc}{2};\frac{ac\sqrt{ac}}{a+c}\le\frac{ac}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=Σ\frac{ab\sqrt{ab}}{a+b}\le\frac{ab+bc+ca}{2}=VP\)
Khi \(a=b=c\)
b)Áp dụng tiếp AM-GM:
\(b\sqrt{a-1}\le\frac{b\left(a-1+1\right)}{2}=\frac{ab}{2}\)
\(a\sqrt{b-1}\le\frac{a\left(b-1+1\right)}{2}=\frac{ab}{2}\)
Cộng theo vế 2 BĐT trên ta có:
\(VT=b\sqrt{a-1}+a\sqrt{b-1}\le ab=VP\)
Khi \(a=b=1\)
30 tháng 4 2017
\(\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow\frac{a+b}{2}-\sqrt{ab}\ge0\)
\(\Leftrightarrow\frac{a+b-2\sqrt{ab}}{2}\ge0\)
\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\ge0\) (luôn đúng)
Vậy \(\frac{a+b}{2}\ge\sqrt{ab}\) (1)
\(\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\)
\(\Leftrightarrow\sqrt{ab}\ge\frac{2ab}{a+b}\)
\(\Leftrightarrow\sqrt{ab}\ge\frac{2\sqrt{ab}^2}{a+b}\)
\(\Leftrightarrow\frac{2\sqrt{ab}}{a+b}\le1\)
\(\Leftrightarrow\frac{2\sqrt{ab}}{a+b}-1\le0\)
\(\Leftrightarrow\frac{2\sqrt{ab}-a-b}{a+b}\le0\)
\(\Leftrightarrow\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{a+b}\le0\) (luôn đúng)
Vậy \(\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a+b}{2}\ge\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\) (đpcm)
24 tháng 9 2020
1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)
2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)
4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2019
\(P=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1-xy}\right):\left(\frac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\left(\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\frac{\left(x+1\right)\left(y+1\right)}{1-xy}\right)\)
\(=\frac{2\sqrt{x}\left(y+1\right)}{\left(1-xy\right)}.\frac{\left(1-xy\right)}{\left(x+1\right)\left(y+1\right)}=\frac{2\sqrt{x}}{x+1}\)
\(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}-1\)
\(\Rightarrow P=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}=\frac{2+6\sqrt{3}}{13}\)
Ta có \(1-P=1-\frac{2\sqrt{x}}{x+1}=\frac{x-2\sqrt{x}+1}{x+1}=\frac{\left(\sqrt{x}-1\right)^2}{x+1}\ge0\) \(\forall x\ge0\)
\(\Rightarrow1-P\ge0\Rightarrow P\le1\)