Cho \(\sqrt{a+3}+\sqrt{b+3}=4\)
Tìm max \(\sqrt{a}+\sqrt{b}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
28 tháng 2 2022
\(\Leftrightarrow P=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)
\(=\dfrac{\sqrt{3\left(a-3\right)}}{a\sqrt{3}}+\dfrac{\sqrt{4\left(b-4\right)}}{2b}+\dfrac{\sqrt{2\left(c-2\right)}}{c\sqrt{2}}\le\dfrac{\dfrac{3+a-3}{2}}{a\sqrt{3}}+\dfrac{\dfrac{4+b-4}{2}}{2b}+\dfrac{\dfrac{2+c-2}{2}}{c\sqrt{2}}=\dfrac{1}{2\sqrt{3}}+\dfrac{1}{4}+\dfrac{1}{2\sqrt{2}}\)
\(dấu"="xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}3=a-3\\4=b-4\\2=c-2\\\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)
29 tháng 11 2021
\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)
\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)
\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)
29 tháng 11 2021
\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)
HM
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2022
\(P=\dfrac{1}{2}\sqrt{4a\left(b+3\right)}+\dfrac{1}{2}\sqrt{4b\left(a+3\right)}\)
\(P\le\dfrac{1}{4}\left(4a+b+3\right)+\dfrac{1}{4}\left(4b+a+3\right)\)
\(P\le\dfrac{1}{4}\left(5a+5b+6\right)\le\dfrac{1}{4}\left(5.2+6\right)=4\)
\(P_{max}=4\) khi \(a=b=1\)
MH
25 tháng 4 2023
Đề thiếu nhé, a,b,c >0
Áp dụng BĐT Bunhiacopxki, ta có:
\(M^2=\left(\sqrt{2a+5\sqrt{ab}+2b}+\sqrt{2b+5\sqrt{bc}+2c}+\sqrt{2c+5\sqrt{ca}+2a}\right)^2\)
\(\le3\left[4\left(a+b+c\right)+5\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\right]\)
\(\le3\left[4\left(a+b+c\right)+5\left(a+b+c\right)\right]=81\)
\(\Rightarrow M\le9\)
\(MaxM=9\Leftrightarrow a=b=c=1\)
(\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\sqrt{\left(a+b+c\right)\left(a+b+c\right)}=a+b+c\left(Bunhiacopxki\right)\))
PM
3
KN
17 tháng 12 2019
Ta có:
\(\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{\left(a+b\right)+\frac{2}{3}+\frac{2}{3}}{3}\)
Tương tự:
\(\sqrt[3]{b+c}\le\frac{\left(b+c\right)+\frac{2}{3}+\frac{2}{3}}{3}\)
\(\sqrt[3]{c+a}\le\frac{\left(c+a\right)+\frac{2}{3}+\frac{2}{3}}{3}\)
\(\Rightarrow\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a}\le\sqrt[3]{\frac{9}{4}}.\frac{2\left(a+b+c\right)+4}{3}\)
\(=\sqrt[3]{\frac{9}{4}}.\frac{6}{3}=\sqrt[3]{18}\)
(Dấu "="\(\Leftrightarrow\hept{\begin{cases}a+b=\frac{2}{3}\\b+c=\frac{2}{3}\\c+a=\frac{2}{3}\end{cases}}\)\(\Leftrightarrow a=b=c=\frac{1}{3}\))
20 tháng 12 2019
Em làm sai tại đây nhé:
\(\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\sqrt[3]{\frac{9}{4}}.\frac{1}{3}.\left(a+b+\frac{2}{3}+\frac{2}{3}\right)\)