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gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
ta có:
\(A^2=\left(\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\right)^2\le\left(a+b+c\right)\left(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\right)\) (BĐT Bu-nhi-a)
=>\(A^2\le\sqrt{3}\left(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\right)\) (*)
mặt khác ta có: \(a^2+1\ge2a\) (BĐT cauchy ) =>\(\frac{a}{a^2+1}\le\frac{1}{2}\)
tương tự ta có: \(\frac{b}{b^2+1}\le\frac{1}{2}\) ; \(\frac{c}{c^2+1}\le\frac{1}{2}\)
=> \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\) (**)
từ (*),(**) => \(A^2\le\sqrt{3}.\frac{3}{2}=\frac{3\sqrt{3}}{2}\)
=>\(A\le\sqrt{\frac{3\sqrt{3}}{2}}\)
=> GTLN của A là \(\sqrt{\frac{3\sqrt{3}}{2}}\) <=> a=b=c<\(\frac{\sqrt{3}}{3}\)
Ta có:
\(\frac{a}{\sqrt{a^2+1}}=\frac{a}{\sqrt{a^2+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}\)
\(\le\frac{\sqrt[8]{27}a}{\sqrt{4\sqrt[4]{a^2}}}=\frac{\sqrt[8]{27a^6}}{2}\)
\(=\frac{\sqrt{3}}{2}.\sqrt[8]{a^6.\frac{1}{3}}\)
\(\le\frac{\sqrt{3}}{2}.\frac{6a+\frac{2}{\sqrt{3}}}{8}\left(1\right)\)
Tương tự ta cũng có:
\(\hept{\begin{cases}\frac{b}{\sqrt{b^2+1}}\le\frac{\sqrt{3}}{2}.\frac{6b+\frac{2}{\sqrt{3}}}{8}\left(2\right)\\\frac{c}{\sqrt{c^2+1}}\le\frac{\sqrt{3}}{2}.\frac{6c+\frac{2}{\sqrt{3}}}{8}\left(3\right)\end{cases}}\)
Từ (1), (2), (3)
\(\Rightarrow A\le\frac{\sqrt{3}}{2}.\left(\frac{6}{8\sqrt{3}}+\frac{6}{8}\left(a+b+c\right)\right)\)
\(\le\frac{\sqrt{3}}{2}.\left(\frac{3}{4\sqrt{3}}+\frac{3\sqrt{3}}{4}\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
\(\Leftrightarrow C=\frac{\left(2+\sqrt{a}\right)^2-\left(2-\sqrt{a}\right)^2+4a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\frac{2\sqrt{a}-\sqrt{a}-3}{\sqrt{a}\left(2-\sqrt{a}\right)}\)
\(\Leftrightarrow C=\frac{2\sqrt{a}+2\sqrt{a}+4a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}.\frac{\left(2-\sqrt{a}\right).\sqrt{a}}{\sqrt{a}-3}=\frac{\left(4\sqrt{a}+4a\right)\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)}\)
b) Để C>0 thì \(\frac{4\left(a-\sqrt{a}\right)\sqrt{a}}{\left(\sqrt{a}+2\right)\sqrt{a}+3}>0hay\left(a-\sqrt{a}\right)>0=>a>1\)
c) bổ sung ý c) tìm a để C=-1
để B=-1
\(\Leftrightarrow\left(4\sqrt{a}+4a\right)\sqrt{a}=-\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)\)
\(\Leftrightarrow4a+4a\sqrt{a}=-a+3\sqrt{a}-2\sqrt{a}+6\)
\(\Leftrightarrow5a+4a\sqrt{a}-\sqrt{a}-6=0=>\orbr{\begin{cases}\sqrt{a}=1\\5\sqrt{a}+4a-1=0\left(zô\right)lý\end{cases}=>a=1}\)
Bài 1:
Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
xong bn áp dụng lên trên lm tiếp
Bài 3:
theo bđt cô si ta có:
\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
Tương tự ta có :
\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
Cộng vế vs vế (1)(2)(3) ta có:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)