Chứng minh rằng: \(a\ge b\ge0\) thì \(\sqrt{a}-\sqrt{b}\le\sqrt{a-b}\)
a)Cho a,b,c \(\ge\)0, a+b+c\(\le\)1.Chứng minh rằng:\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
b)Cho a,b,c \(\ge\)0, a+b+c\(\le\)6.Chứng minh rằng: \(\sqrt{a+\sqrt{b+\sqrt{2c}}}+\sqrt{b+\sqrt{c+\sqrt{2a}}}+\sqrt{c+\sqrt{a+\sqrt{2b}}}\le6\)
a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
\(\le2\cdot\left(1+1+1\right)\left(a+b+c\right)\le6\)
\(\Rightarrow VT^2\le6\Rightarrow VT\le\sqrt{6}=VP\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
b)Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{a+\sqrt{b+\sqrt{2c}}}+\sqrt{b+\sqrt{c+\sqrt{2a}}}+\sqrt{c+\sqrt{a+\sqrt{2b}}}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+c+Σ\sqrt{b+\sqrt{2c}}\right)\)
\(=3\left(6+\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)\)
Đặt \(A^2=\left(\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+c+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\)
\(=3\left(6+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\)
Đặt tiếp: \(B^2=\left(\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)^2\)
\(\le2\cdot\left(1+1+1\right)\left(a+b+c\right)\le36\Rightarrow B\le6\)
\(\Rightarrow A^2\le3\left(6+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\le3\cdot12=36\Rightarrow A\le6\)
\(\Rightarrow VT^2\le3\left(6+\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)\)
\(\le3\left(6+6\right)=3\cdot12=36\Rightarrow VT\le6=VP\)
Xảy ra khi \(a=b=c=2\)
Chứng minh đẳng thức sau với \(b\ge0;a\ge\sqrt{b}\)
\(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\)
\(\Rightarrow\left(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}\right)^2=\left(\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\right)^2\Leftrightarrow a+\sqrt{b}+a-\sqrt{b}\mp2\sqrt{\left(a+\sqrt{b}\right)\cdot\left(a-\sqrt{b}\right)}=2a\mp2\sqrt{a^2-b}\Leftrightarrow2a\mp2\sqrt{a^2-b}=2a\mp2\sqrt{a^2-b}\) (luôn đúng) \(\Rightarrowđpcm\)
Giải giùm mình mấy bài BPT này nha
a) Chứng minh: \(\dfrac{a+b}{2}\le\sqrt{\dfrac{a^2+b^2}{2}}\)
b) Cho a,b>0 chứng minh: \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)
c) Cho a+b\(\ge\)0 chứng minh: \(\dfrac{a+b}{2}\ge\sqrt[3]{\dfrac{a^3+b^3}{2}}\)
d) Chứng minh: \(\dfrac{a+b+c}{3}\ge\sqrt{\dfrac{ab+bc+ac}{3}}\) ; \(a,b,c\ge0\)
e) Chứng minh: \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
Chứng minh rằng:
a, \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
b, \(\sqrt{a}-\sqrt{b}\le\sqrt{a-b}\)
a> ĐK a,b>=0
ta có : \(\sqrt{ab}\ge0\Rightarrow2\sqrt{ab}\ge0\Rightarrow a+b+2\sqrt{ab}\ge a+b\Rightarrowđpcm\)
* xét b>a => đpcm
*xét a>b bạn dùng cm tương đương
bình phương 2 vế rồi trừ 2 vế cho nhau thì nó ra 1 đg thức luôn đúng
Cho \(a,b,c\ge0\)
Chứng minh rằng: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Ta có ; \(\hept{\begin{cases}\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\\\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\\\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b\ge2\sqrt{ab}\\b+c\ge2\sqrt{bc}\\c+a\ge2\sqrt{ac}\end{cases}}\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
Câu 1: Rút gọn biểu thức
a) \(N=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
b) \(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
Câu 2:
a) Cho a > 0. Chứng minh: \(a+\dfrac{1}{a}\ge2\)
b) Cho \(a\ge0\) , \(b\ge0\) . Chứng minh: \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
c) Cho a, b > 0. Chứng minh: \(\sqrt{a}+\sqrt{b}\le\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\)
d) Chứng minh: \(\dfrac{a^2+2}{\sqrt{a^2+1}}\ge2\) với mọi a
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
2b)
Biến đổi tương đương:
\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\) (1)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\dfrac{a+2\sqrt{ab}+b}{4}\)
\(\Leftrightarrow2a+2b\ge a+2\sqrt{ab}+b\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) luôn đúng
=> (1) đúng
Dấu "=" xảy ra khi a = b.
2c)
Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:
\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\) (đpcm)
Dấu "=" xảy ra khi a = b.
2d)
Áp dụng BĐT AM - GM, ta có:
\(\dfrac{a^2+2}{\sqrt{a^2+1}}=\dfrac{a^2+1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\dfrac{1}{\sqrt{a^2+1}}\ge2\) (đpcm)
Dấu "=" xảy ra khi a = 0
cho a,b, c\(\ge\)0; a+b+c=1. Chứng minh rằng\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
Áp dụng BĐT Bunhiacopxki, ta có :
\(\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\)
\(\Rightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le6\left(a+b+c\right)\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
1. Cho \(a\ge0;b\ge0.\) Chứng minh: \(\left(a+1\right)\sqrt{b}+\left(b+1\right)\sqrt{a}\le\left(a+1\right)\left(b+1\right)\)
2. Cho \(a\ge2;b\ge\frac{1}{2}\)Chứng minh: \(a\sqrt{2b-1}+2b\sqrt{2a-4}\le2ab\)
Cho \(a,b,c\ge0\), Chứng minh rằng \(a^3+b^3+c^3\ge a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\)