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NC
13 tháng 8 2020
Bài làm:
Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
=> đpcm
13 tháng 8 2020
Ta có: \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-\left(a-2\sqrt{ab}+b\right)+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4b+4\sqrt{ab}}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(đpcm)
UK
10 tháng 11 2017
Cái này không khó :v
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+c}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Face khác ;v, theo AM-GM, ta có
\(\dfrac{a+b+c}{2}\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{6}{2}=3\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c=2
10 tháng 11 2017
Ta có :
\(\frac{a^2}{a+b}=\frac{a^2+ab-ab}{a+b}=a-\frac{ab}{a+b}\le a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)(1)
Tương tự \(\hept{\begin{cases}\frac{b^2}{b+c}\le b-\frac{\sqrt{bc}}{2}\\\frac{c^2}{a+c}\le c-\frac{\sqrt{ac}}{2}\end{cases}}\)(2)
Nhhan (1);(2) lại ta được
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge a+b+c-\frac{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}}{2}=a+b+c-3\)
Ta lại có : \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{bc}=6\) (tự cm)
\(\Rightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge6-3=3\)(đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2019
a/
\(=\frac{a+b}{b^2}.\frac{\left|a\right|.b^2}{\left|a+b\right|}=\frac{\left(a+b\right).b^2.\left|a\right|}{b^2\left(a+b\right)}=\left|a\right|\)
b/
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{2\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
6 tháng 7 2019
BĐT<=>
\(\left(\frac{2ab}{a+b}-\frac{a+b}{2}\right)+\left(\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\right)\ge0\)
<=> \(-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\frac{a^2+b^2}{2}-ab}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}\ge0\)
<=> \(\frac{\left(a-b\right)^2}{2(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab})}-\frac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)
<=> \(a+b\ge\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\)
<=> \(\frac{a^2+b^2}{2}+ab\ge2\sqrt{\frac{a^2+b^2}{2}.ab}\)luôn đúng
=> ĐPCM
Dấu bằng xảy ra khi a=b
NY
3 tháng 2 2020
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
PG
3 tháng 2 2020
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
\(\frac{a^2+b^2}{ab}+\frac{\sqrt{2ab}}{a+b}=\frac{\left(a+b\right)^2}{ab}+\frac{\sqrt{2ab}}{a+b}-2\)
đặt \(t=\frac{a+b}{\sqrt{ab}}\left(t\ge2\right)\)(do \(a+b\ge2\sqrt{ab}\)
\(A=t^2+\frac{2}{t}-2=\left(\frac{1}{t}+\frac{1}{t}+\frac{t^2}{8}\right)+\frac{7}{8}t^2-2\ge3\sqrt[3]{\frac{1.1.t^2}{t.t.8}}+\frac{7}{8}.2^2-2=3\)
vậy ..................
con láo