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 \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\)
1/ Cho a,b>0 , thỏa mãn ab = 1. Chứng minh rằng:
\(\dfrac{a}{\sqrt{b+2}}+\dfrac{b}{\sqrt{a+2}}+\dfrac{1}{\sqrt{a+b+ab}}\ge\sqrt{3}\)
2/ Cho a>0. Chứng minh rằng:
a+\(\dfrac{1}{a}\ge\sqrt{\dfrac{1}{a^2+1}}+\sqrt{1+\dfrac{1}{a^2+1}}\)
3/ Cho a, b>0. Chứng minh rằng:
2(a+b)\(\le1+\sqrt{1+4\left(a^3+b^3\right)}\)
Cho a, b, c không âm thỏa mãn a + b + c = 3
a. Chứng minh rằng \(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\ge\sqrt{a^2+b^2+c^2+15}\)
b. Chứng minh rằng \(\sum\dfrac{a+1}{a^2+2a+3}\le1\)
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: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)
a)Cho a>b>0 chứng minh rằng \(\frac{1}{a+b}\le\frac{1}{2\sqrt{ab}}\)
b) Chứng minh \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+\frac{\sqrt{4}-\sqrt{3}}{7}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}< \frac{1}{2}\)
* Cho a, b, c ≥ 0. Chứng minh rằng a+b+c ≥ \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
* Chứng minh rằng A=\(\sqrt{1+2008^2+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}\)có giá trị là số tự nhiên
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 \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)
a)cho a>b>0 chứng minh rằng : \(\dfrac{1}{a+b}\le\dfrac{1}{2\sqrt{ab}}\)
b) Chứng minh \(\dfrac{\sqrt{2}-\sqrt{1}}{3}+\dfrac{\sqrt{3}-\sqrt{2}}{5}+\dfrac{\sqrt{4}-\sqrt{3}}{7}+...+\dfrac{\sqrt{2011}-\sqrt{2010}}{4021}< \dfrac{1}{2}\)
giúp mk vs
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)
Bài 3:
a) cho a≥1,b≥1. Chứng minh: a\(\sqrt{b-1}\)+b\(\sqrt{a-1}\) ≤ ab
b) ) Cho 4 số thực dương a, b, c, d. Chứng minh rằng: \(\sqrt{ac}+\sqrt{bd}\)≤\(\sqrt{\left(a+b\right)\left(c+d\right)}\)
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 \(\dfrac{\sqrt{a}}{\sqrt{c}}=\dfrac{\sqrt{b}}{\sqrt{d}}\Leftrightarrow ad=bc\)
\(b,\) 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 \("="\Leftrightarrow ad=bc\)
Cho a,b,c > 0 và ab + bc + ac = 1. Chứng minh rằng :\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) 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 \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Cho \(a,b>0\); \(c< 0\). Chứng minh rằng:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
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
Cho ba số dương a,b,c thỏa mãn abc = 1. Chứng minh rằng :
\(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\) ≤ \(\dfrac{1}{2}\)