a/Cho a>0 Chứng minh rằng:\(\sqrt{a}+1\)>\(\sqrt{a+1}\)
b/Cho a>1 Chứng minh rằng \(\sqrt{a-1}\)<\(\sqrt{a}\)
c/Chứng minh rằng \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\)
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
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 3 2023
a.
Bình phương 2 vế, BĐT cần chứng minh trở thành:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)
Ta có:
\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cộng lại:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
b.
\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)
Nên ta chỉ cần chứng minh:
\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)
\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)
Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)
26 tháng 9 2021
Bài 1:
Ta có: \(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(a+c\ge2\sqrt{ac}\)
Do đó: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)
hay \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2018
Lời giải:
a) Ta thấy: \(a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0, \forall a,b>0\)
\(\Rightarrow a+b\geq 2\sqrt{ab}>0\Rightarrow \frac{1}{a+b}\le \frac{1}{2\sqrt{ab}}\).
Vì $a> b$ nên dấu bằng không xảy ra . Tức \(\frac{1}{a+b}< \frac{1}{2\sqrt{ab}}\)
Ta có đpcm
b)
Áp dụng kết quả phần a:
\(\frac{1}{3}=\frac{1}{1+2}< \frac{1}{2\sqrt{2.1}}\)
\(\frac{1}{5}=\frac{1}{3+2}< \frac{1}{2\sqrt{2.3}}\)
\(\frac{1}{7}=\frac{1}{4+3}< \frac{1}{2\sqrt{4.3}}\)
.....
\(\frac{1}{4021}=\frac{1}{2011+2010}< \frac{1}{2\sqrt{2011.2010}}\)
Do đó:
\(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{\sqrt{2}-\sqrt{1}}{2\sqrt{2.1}}+\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3.2}}+\frac{\sqrt{4}-\sqrt{3}}{2\sqrt{4.3}}+....+\frac{\sqrt{2011}-\sqrt{2010}}{2\sqrt{2011.2010}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2010}}-\frac{1}{2\sqrt{2011}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{2011}}< \frac{1}{2}\) (đpcm)
22 tháng 9 2021
a)Áp dụng AM-GM có:
\(a\sqrt{b-1}\le a.\dfrac{b-1+1}{2}=\dfrac{ab}{2}\)
\(b\sqrt{a-1}\le b.\dfrac{a-1+1}{2}=\dfrac{ab}{2}\)
\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le\dfrac{ab}{2}+\dfrac{ab}{2}\)
\(\Leftrightarrow a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
Dấu "=" xảy ra khi a=b=2
b)Áp dụng bđt bunhiacopxki có:
\(\left(\sqrt{ac}+\sqrt{bd}\right)^2=\left(\sqrt{a}.\sqrt{c}+\sqrt{b}.\sqrt{d}\right)^2\)\(\le\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]=\left(a+b\right)\left(c+d\right)\)
\(\Rightarrow\sqrt{ac}+\sqrt{bd}\le\sqrt{\left(a+b\right)\left(c+d\right)}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{a}}{\sqrt{c}}=\dfrac{\sqrt{b}}{\sqrt{d}}\Leftrightarrow ad=bc\)
22 tháng 9 2021
\(b,\) Áp dụng BĐT Bunhiacopski:
\(\left(a+b\right)\left(c+d\right)=\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]\\ \ge\left(\sqrt{ac}+\sqrt{bd}\right)^2\)
Dấu \("="\Leftrightarrow ad=bc\)
6 tháng 3 2021
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 3 2021
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
22 tháng 11 2021
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
Cần cm:
\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\\ \Leftrightarrow a+b=a+b+2c+2\sqrt{\left(a+c\right)\left(b+c\right)}\\ \Leftrightarrow2c+2\sqrt{ab+ac+bc+c^2}=0\\ \Leftrightarrow2c+2\sqrt{c^2}=0\\ \Leftrightarrow2c+2\left|c\right|=0\\ \Leftrightarrow2c-2c=0\left(c< 0\right)\\ \Leftrightarrow0=0\left(luôn.đúng\right)\)
Vậy đẳng thức đc cm
a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) )
b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)
c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm )