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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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
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:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
Áp dụng BĐT Cauchy-Schwarz:
$\frac{a}{a+\sqrt{2016a + bc}}=\frac{a}{a+\sqrt{(a+b+c)a + bc}} =\frac{a}{a+\sqrt{(a+b)(c+a)}} \leq \frac{a}{a+\sqrt{(\sqrt{ab}+\sqrt{ac})^{2}}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$
$\Rightarrow \frac{a}{a+\sqrt{2016a + bc}} + \frac{b}{b+\sqrt{2016b + ca}} + \frac{c}{c+\sqrt{2016c + ab}}\leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$
...............................
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Leftrightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Áp dụng BĐT AM-GM:
\(\frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}(\frac{b}{a+b}+\frac{c}{a+c})\)
Hoàn toàn tương tự:
\(\frac{1}{\sqrt{b^2+1}}=\sqrt{\frac{ac}{(b+a)(b+c)}}\leq \frac{1}{2}(\frac{a}{b+a}+\frac{c}{b+c})\)
\(\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}(\frac{a}{c+a}+\frac{b}{b+c})\)
Cộng theo vế:
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\leq \frac{1}{2}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)
\(\Leftrightarrow2n+1=1\left(2n+1\right)\)
\(\Leftrightarrow2n+1=2n+1\)
\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
Câu b) ý 2:
Áp dụng BĐT cô si ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
nhầm mọi người ơi chứng minh cho mình <=\(\dfrac{3}{\sqrt{2}}\)