Cho a+b+c+ab+bc+ac+abc=7. CMR: \(M=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge3\sqrt{2}\)
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23 tháng 6 2021
Áp dụng bđt cosi schwart ta có:
`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`
Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`
`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`
Dấu "=" `<=>a=b=c=1.`
AH
Akai Haruma
Giáo viên
30 tháng 11 2018
Lời giải:
Áp dụng BĐT AM-GM (Cô-si)
\(1+a^3+b^3\geq 3\sqrt[3]{a^3b^3}=3ab\)
\(\Rightarrow \frac{\sqrt{1+a^3+b^3}}{ab}\geq \frac{\sqrt{3ab}}{ab}=\frac{c\sqrt{3ab}}{abc}=c\sqrt{3ab}=\sqrt{c}.\sqrt{3abc}=\sqrt{3c}\)
Hoàn toàn tương tự:
\(\frac{\sqrt{1+b^3+c^3}}{bc}\geq \sqrt{3a}\)
\(\frac{\sqrt{1+a^3+c^3}}{ac}\geq \sqrt{3b}\)
Cộng theo vế những BĐT vừa thu được ta có:
\(\frac{\sqrt{a^3+b^3+1}}{ab}+\frac{\sqrt{b^3+c^3+1}}{bc}+\frac{\sqrt{c^3+a^3+1}}{ac}\geq \sqrt{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})\)
\(\geq \sqrt{3}.3\sqrt[3]{\sqrt{a}.\sqrt{b}.\sqrt{c}}=\sqrt{3}.3\sqrt[6]{abc}=3\sqrt{3}\) (áp dụng BĐT Cô-si)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
UK
10 tháng 11 2017
Cái này không khó :v
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+c}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Face khác ;v, theo AM-GM, ta có
\(\dfrac{a+b+c}{2}\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{6}{2}=3\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c=2
10 tháng 11 2017
Ta có :
\(\frac{a^2}{a+b}=\frac{a^2+ab-ab}{a+b}=a-\frac{ab}{a+b}\le a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)(1)
Tương tự \(\hept{\begin{cases}\frac{b^2}{b+c}\le b-\frac{\sqrt{bc}}{2}\\\frac{c^2}{a+c}\le c-\frac{\sqrt{ac}}{2}\end{cases}}\)(2)
Nhhan (1);(2) lại ta được
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge a+b+c-\frac{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}}{2}=a+b+c-3\)
Ta lại có : \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{bc}=6\) (tự cm)
\(\Rightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge6-3=3\)(đpcm)
MD
22 tháng 4 2018
\(\Leftrightarrow y=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)
Ta có: \(\dfrac{\sqrt{c-2}}{c}\le\dfrac{1}{2\sqrt{2}}\Leftrightarrow\left(\sqrt{c-2}-\sqrt{2}\right)^2\ge0\) ( Luôn đúng)
Tương tự: \(\dfrac{\sqrt{a-3}}{a}\le\dfrac{1}{2\sqrt{3}};\dfrac{\sqrt{b-4}}{b}\le\dfrac{1}{4}\)
\(\Rightarrow y\le\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{4}\) và dấu ''='' xảy ra khi c = 4; a = 6; b = 8
N
10 tháng 9 2017
set \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\)\(\Rightarrow x+y+z=3\)
\(VT=\sum\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)}{4x}}=\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}.\left(\sum\dfrac{1}{\sqrt{4x\left(y+z\right)}}\right)\)
Áp dụng BĐT AM-GM:
\(\dfrac{1}{\sqrt{4x\left(y+z\right)}}+\dfrac{1}{\sqrt{4y\left(x+z\right)}}+\dfrac{1}{\sqrt{4z\left(x+y\right)}}\ge\dfrac{9}{2\left(\sqrt{xy+xz}+\sqrt{yz+yx}+\sqrt{xz+zy}\right)}\)
Áp dụng BĐT bunyakovsky:
\(\sum\sqrt{xy+yz}\le\sqrt{6\left(xy+yz+xz\right)}\)
\(\Rightarrow\sum\dfrac{1}{2\sqrt{x\left(y+z\right)}}\ge\dfrac{9}{2\sqrt{6\left(xy+yz+xz\right)}}\)
Mà \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{3}\left(xy+yz+xz\right)\)(*)
\(\Rightarrow VT\ge\sqrt{\dfrac{8}{3}\left(xy+yz+xz\right)}.\dfrac{9}{2\sqrt{6\left(xy+yz+xz\right)}}=3\)
Dấu = xảy ra khi x=y=z hay a=b=c=1
(*) Prove BĐT \(\left(m+n\right)\left(n+p\right)\left(m+p\right)\ge\dfrac{8}{9}\left(m+n+p\right)\left(mn+np+pm\right)\)
khai triển ,để ý rằng \(\left(m+n\right)\left(n+p\right)\left(p+m\right)=\left(m+n+p\right)\left(mn+np+pm\right)-mnp\)
4 tháng 2 2021
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12 tháng 7 2016
Ta có : \(\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ac\sqrt{b-4}}{abc}=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)
Áp dụng bất đẳng thức Cauchy, ta có :
\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\le\frac{2+c-2}{2\sqrt{2}c}=\frac{1}{2\sqrt{2}}\)
\(\frac{\sqrt{a-3}}{a}=\frac{\sqrt{3\left(a-3\right)}}{\sqrt{3}a}\le\frac{3+a-3}{2\sqrt{3}a}=\frac{1}{2\sqrt{3}}\)
\(\frac{\sqrt{b-4}}{b}=\frac{\sqrt{4\left(b-4\right)}}{2b}\le\frac{4+b-4}{4b}=\frac{1}{4}\)
\(\Rightarrow\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\le\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}c-2=2\\b-4=4\\a-3=3\end{cases}\Leftrightarrow}\hept{\begin{cases}c=4\\b=8\\a=6\end{cases}}\)
Vậy giá trị lớn nhất của biểu thức là \(\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\Leftrightarrow\hept{\begin{cases}a=6\\b=8\\c=4\end{cases}}\)
SL
12 tháng 7 2016
phá ra nha
sau đó bạn lm theo tek này
\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\le\frac{\frac{c}{2}}{\sqrt{2}c}=\frac{1}{\sqrt{2}}\)
mấy cái kia tt nha