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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)
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)=> \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=\frac{a+b+c}{2}\left(đpcm\right)\)
Bài 1 :
Áp dụng bất đẳng thức Cauchy ta có :
\(\frac{\left(x-1\right)^2}{z}+\frac{z}{4}\ge2\sqrt{\frac{\left(x-1\right)^2}{z}\frac{z}{4}}=\left|x-1\right|=1-x\)
\(\frac{\left(y-1\right)^2}{x}+\frac{x}{4}\ge2\sqrt{\frac{\left(y-1\right)^2}{x}\frac{x}{4}}=\left|y-1\right|=1-y\)
\(\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge2\sqrt{\frac{\left(z-1\right)^2}{y}\frac{y}{4}}=\left|z-1\right|=1-z\)
\(\Rightarrow\frac{\left(x-1\right)^2}{z}+\frac{z}{4}+\frac{\left(y-1\right)^2}{x}+\frac{x}{4}+\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge1-x+1-y+1-z\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge3-\left(x+y+z\right)-\frac{x+y+z}{4}=3-2-\frac{2}{4}=\frac{1}{2}\)
Vậy GTNN của \(A=\frac{1}{2}\Leftrightarrow x=y=z=\frac{2}{3}\)
a) Áp dụng bdt cosi schwars ta có
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\frac{a+b+c+d}{2}\)
bh mk can mn ho tro jup mk 2 cau cuoi nha
a) Áp dụng bđt Bunhiacopxki ta có :
\(\left[\left(\frac{a}{\sqrt{a+b}}^2\right)+\left(\frac{b}{\sqrt{b+c}}\right)^2+\left(\frac{c}{\sqrt{c+d}}\right)^2+\left(\frac{d}{\sqrt{d+a}}\right)\right]\)\(\times\)\(\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+d}\right)^2+\left(\sqrt{d+a}\right)^2\right]\)\(\ge\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\right)\times2\left(a+b+c+d\right)\ge\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{a+b+c+d}{2}\)( chia cả 2 vế cho \(2\left(a+b+c+d\right)\))
Dấu "=" xảy ra khi : a = b = c = d
Vậy ...
b) Áp dụng bđt Cauchy cho 2 số không âm ta có :
\(\sqrt{a+2}\le\frac{a+2+1}{2}\)
\(\sqrt{b+2}\le\frac{b+2+1}{2}\)
\(\sqrt{c+2}\le\frac{c+2+1}{2}\)
\(\Rightarrow\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\le\frac{a+b+c+2+1+2+1+2+1}{2}\)
Mà \(a+b+c=1\)
\(\Rightarrow\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\le5\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}a+2=1\\b+2=1\\c+2=1\end{cases}}\) \(\Leftrightarrow a=b=c=-1\)
Mà \(a;b;c\ge0\)
Nên dấu "=" không xảy ra
Vậy \(\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}< 5\)
c) mạn phép sửa đề :) \(\left(a,b,c,d\ge0;a+b+c=1\right)\)
Áp dụng bđt Cauchy cho 2 số không âm ta có :
\(\sqrt{a+3}\le\frac{a+3+1}{2}\)
\(\sqrt{b+3}\le\frac{b+3+1}{2}\)
\(\sqrt{c+3}\le\frac{c+3+1}{2}\)
\(\Rightarrow\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\le\frac{a+b+c+12}{2}\)
Mà \(a+b+c=1\)
\(\Rightarrow VT\le6,5\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}c+3=1\\b+3=1\\c+3=1\end{cases}}\) \(\Leftrightarrow a=b=c=-2\)
Mà \(a+b+c=1\)
Nên dấu "=" không xảy ra
Vậy ...
d) Áp dụng bđt Bunhiacopxki ta có :
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\)\(\le\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+d}\right)^2+\left(\sqrt{d+a}\right)^2\right]\)\(\left(1+1+1+1\right)\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\le\left(a+b+b+c+c+d+d+a\right)\times4\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\le2\left(a+b+c+d\right)\times4\)
Mà \(a+b+c+d=1\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\le8\)
\(\Leftrightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\le\sqrt{8}\)
Dấu "=" xảy ra khi : \(a=b=c=d\)
Vậy ...