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Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
Lời giải:
Áp dụng BĐT AM-GM ngược dấu ta có:
\(A=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ac}}=\frac{ab}{\sqrt{c(a+b+c)+ab}}+\frac{bc}{\sqrt{a(a+b+c)+bc}}+\frac{ca}{\sqrt{b(a+b+c)+ac}}\)
\(=\frac{ab}{\sqrt{(c+a)(c+b)}}+\frac{bc}{\sqrt{(a+b)(a+c)}}+\frac{ca}{\sqrt{(b+a)(b+c)}}\)
\(\leq \frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}\right)+\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)+\frac{1}{2}\left(\frac{ca}{b+a}+\frac{ca}{b+c}\right)\)
\(A\leq \frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}(b+a+c)=\frac{1}{2}\)
Vậy \(A_{\max}=\frac{1}{2}\) tại \(a=b=c=\frac{1}{3}\)
Ta có
\(\sum\dfrac{a}{a+\sqrt{2019a+bc}}=\sum\dfrac{a}{a+\sqrt{a^2+a\left(b+c\right)+bc}}\)
Áp dụng AM - GM : \(b+c\ge2\sqrt{bc}\)
\(\Rightarrow\sum\dfrac{a}{a+\sqrt{a^2+a\left(b+c\right)+bc}}\le\dfrac{a}{a+\sqrt{a^2+2a\sqrt{bc}+bc}}\)
\(=\sum\dfrac{a}{a+\sqrt{\left(a+\sqrt{bc}\right)^2}}=\sum\dfrac{a}{a+a+\sqrt{bc}}\)
Tự làm tiếp
Ta có:
\(\dfrac{ab}{\sqrt{c+ab}}=\dfrac{ab}{\sqrt{c\left(a+b+c\right)+ab}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=\dfrac{\sqrt{ab}}{\sqrt{a+c}}.\dfrac{\sqrt{ab}}{\sqrt{b+c}}\)
\(\Rightarrow\dfrac{ab}{\sqrt{c+ab}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự ta có:
\(\dfrac{bc}{\sqrt{a+bc}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\) ; \(\dfrac{ac}{\sqrt{b+ac}}\le\dfrac{1}{2}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)\)
Cộng vế với vế ta được:
\(A\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}\right)\)
\(\Rightarrow A\le\dfrac{1}{2}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right)\)
\(\Rightarrow A\le\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
\(\Rightarrow A_{max}=\dfrac{1}{2}\) khi \(a=b=c=\dfrac{1}{3}\)
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
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 :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
\(T=\dfrac{a+b}{\sqrt{ab+c}}+\dfrac{b+c}{\sqrt{bc+a}}+\dfrac{c+a}{\sqrt{ca+b}}\)
\(\odot\) Ta có: \(\dfrac{a+b}{\sqrt{ab+c}}=\dfrac{a+b}{\sqrt{ab+c\left(a+b+c\right)}}=\dfrac{a+b}{\sqrt{\left(b+c\right)\left(a+c\right)}}\)
\(\odot\) Tương tự:
\(\dfrac{b+c}{\sqrt{bc+a}}=\dfrac{b+c}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\dfrac{c+a}{\sqrt{ca+b}}=\dfrac{c+a}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\odot\) Áp dụng bất đẳng thức AM - GM
\(\Rightarrow T=\dfrac{a+b}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{b+c}{\sqrt{\left(a+c\right)\left(b+a\right)}}+\dfrac{a+c}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\ge3\sqrt[3]{\dfrac{a+b}{\sqrt{\left(a+c\right)\left(b+c\right)}}\times\dfrac{b+c}{\sqrt{\left(a+c\right)\left(b+a\right)}}\times\dfrac{a+c}{\sqrt{\left(a+b\right)\left(b+c\right)}}}\)
\(=3\)
\(\odot\) Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)