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\(\sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\sqrt{\dfrac{c}{1-c}}>2\)
\(\Leftrightarrow\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế ta cũng có:
\(VT\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu "=" ko xảy ra nên ta có ĐPCM
bạn có thể giải kỹ hơn phần bđt am-gm ko tại sao lại ra lớn hơn luôn 2a/(a+b+c)
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 :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Cho a,b,c dương. Chứng minh: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Áp dụng bđt AM-GM:
\(\left\{{}\begin{matrix}\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\\\sqrt{b\left(c+a\right)}\le\dfrac{a+b+c}{2}\\\sqrt{c\left(a+b\right)}\le\dfrac{a+b+c}{2}\end{matrix}\right.\)
Ta có: \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)
\(=\dfrac{a}{\sqrt{a\left(b+c\right)}}+\dfrac{b}{\sqrt{b\left(a+c\right)}}+\dfrac{c}{\sqrt{c\left(a+b\right)}}\)
\(\ge\dfrac{a}{\dfrac{a+b+c}{2}}+\dfrac{b}{\dfrac{a+b+c}{2}}+\dfrac{c}{\dfrac{a+b+c}{2}}\)
\(=\dfrac{2\left(a+b+c\right)}{\left(a+b+c\right)}=2\)
Dấu "=" ko xảy ra nên ta có đpcm
Lời giải:
Theo hệ quả quen thuộc của BĐT AM-GM thì:
\((a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Leftrightarrow (\sqrt{3})^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq 1\)
\(\Rightarrow \frac{a}{\sqrt{a^2+1}}\leq \frac{a}{\sqrt{a^2+ab+bc+ac}}=\frac{a}{\sqrt{(a+b)(a+c)}}\)
Hoàn toàn TT với các phân thức còn lại và cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\)
\(\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+c}+\frac{b}{b+a}\right)+\frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\) (BĐT Cauchy)
hay \(\text{VT}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)(đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
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:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
Lời giải:
Đặt \(\left ( \sqrt{\frac{a}{b+c}},\sqrt{\frac{b}{a+c}},\sqrt{\frac{c}{a+b}} \right )=(x,y,z)\)
\(\Rightarrow \left\{\begin{matrix} x^2=\frac{a}{b+c}\\ y^2=\frac{b}{a+c}\\ z^2=\frac{c}{a+b}\end{matrix}\right.\Rightarrow \frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=2\)
\(\Leftrightarrow (1-\frac{1}{x^2+1})+(1-\frac{1}{y^2+1})+(1-\frac{1}{z^2+1})=1\)
\(\Leftrightarrow \frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1}=1\)
BĐT cần chứng minh tương đương:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 2(x+y+z)(\star)\)
Áp dụng BĐT Bunhiacopxky:
\(\left ( \frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1} \right )(x^2+1+y^2+1+z^2+1)\geq (x+y+z)^2\)
\(\Leftrightarrow x^2+1+y^2+1+z^2+1\geq (x+y+z)^2\)
\(\Leftrightarrow xy+yz+xz\leq \frac{3}{2}\)
Kết hợp với hệ quả của BĐT AM-GM :
\((xy+yz+xz)^2\geq 3xyz(x+y+z)\)
\(\Rightarrow xy+yz+xz\geq \frac{3xyz(x+y+z)}{xy+yz+xz}\geq \frac{3xyz(x+y+z)}{\frac{3}2{}}=2xyz(x+y+z)\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{2xyz(x+y+z)}{xyz}=2(x+y+z)\)
Do đó BĐT \((\star)\) được chứng minh.
Bài toán hoàn thành. Dấu bằng xảy ra khi \(a=b=c\)
