cho a,b,c\(\ge\)0. a+b+c=1. cm
a)\(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}< 3.5\)
b)\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
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 \(a=b=c=2\)
Cho a, b, c \(\ge\)0; a + b + c = 1. CMR: \(\sqrt[]{a+b}+\sqrt{b+c}+\sqrt{c+1}\le\sqrt{6}\)
Áp dụng BĐT Bunhiakovski
\(VT^2=\left(\sqrt{a+b}.1+\sqrt{b+c}.1+\sqrt{c+a}.1\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)\)
\(=3.2\left(a+b+c\right)=6\)
Do đó \(VT\le\sqrt{6}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{a+b}}{1}=\dfrac{\sqrt{b+c}}{1}=\dfrac{\sqrt{c+a}}{1}\\a+b+c=1\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
cho a,b,c\(\ge\)0; a+b+c=1. Chứng minh \(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
Áp dụng BĐT Bunhiacopxki, ta có :
\((\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=3.2=6\)
\(\Leftrightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
Dấu "=" xảy ra khi và chỉ khi a+b=b+c=c+a => a=b=c =1/3
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 :
\(\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\)
\(\Rightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le6\left(a+b+c\right)\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
Cho a, b, c > 1 và \(\sqrt{a-1}\) + \(\sqrt{b-1}\) + \(\sqrt{c-1}\) \(\le\)\(\dfrac{3}{2}\)
Chứng minh rằng:
\(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}\)
Ta có \(\sqrt{a-1}+\dfrac{1}{\sqrt{a-1}}\) \(=\sqrt{a-1}+\dfrac{1}{4\sqrt{a-1}}+\dfrac{3}{4\sqrt{a-1}}\) \(\ge2\sqrt{\sqrt{a-1}.\dfrac{1}{4\sqrt{a-1}}}+\dfrac{3}{4\sqrt{a-1}}\) \(=1+\dfrac{3}{4\sqrt{a-1}}\).
Lập 2 BĐT tương tự rồi cộng vế theo vế, ta có
\(VT\ge3+\dfrac{3}{4}\left(\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\right)\)
\(\ge3+\dfrac{3}{4}.\dfrac{9}{\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}}\)
\(\ge3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}\) \(=\dfrac{15}{2}\).
ĐTXR \(\Leftrightarrow a=b=c=\dfrac{5}{4}\). Ta có đpcm
Có \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}-\left(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\right)\ge6\) (1)
Ta chứng minh (1) đúng
Áp dụng bất đẳng thức Schwarz :
\(\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}}\ge\dfrac{9}{\dfrac{3}{2}}=6\)Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{a-1}=\sqrt{b-1}=\sqrt{c-1}\\\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\dfrac{5}{4}\)(tm)
b1 cho a,b,c ko âm cmr
a)a+b+c\(\ge\)\(\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)\)
b)a+b+c+d+e\(\ge\)\(\sqrt{a}\left(\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e}\right)\)
c)a+b+1\(\ge\)\(\sqrt{ab}+\sqrt{a}+\sqrt{b}\)
d)a+\(\sqrt{2a}+2\)>0
b2 sử dụng cô-si hoặc bu-nhia-cốp-xki
cho a,b,c thoả mãn a+b+c=1 cmr
a)\(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le3,5\)
b)\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
b3CMR
a)\(19>1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}>18\)
b)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
bạn nào giải giúp mk vs 3 hm nx mk phải nộp r bạn nào giải dc con nào thì giải nhé thanks
Cho a,b,c là số dương. CMR:
1. \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
2. \(a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\le a^3+b^3+c^3\)
3. \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$
$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$
Cộng theo vế và thu gọn:
$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$
Ta có đpcm.
Bài 2:
$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$
$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$
$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$
Cộng theo vế và rút gọn thu được:
$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Bài 3:
Áp dụng BĐT Cauchy-Schwarz:
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
cho a,b,c thỏa \(\left\{{}\begin{matrix}a,b,c>0\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\) chứng minh rằng\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge\sqrt{abc}+\sqrt{a}+\sqrt{b}+\sqrt{\sqrt{c}}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có: \(\sqrt{a+bc}=\sqrt{\dfrac{a^2+abc}{a}}=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a}}\)
thiết lập tương tự ,bất đẳng thức cần chứng minh tương đương:
\(\Leftrightarrow\sum\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a}}\ge\sqrt{abc}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(\Leftrightarrow\sum\sqrt{bc\left(a+b\right)\left(a+c\right)}\ge abc+\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\Leftrightarrow\sum\sqrt{\left(b^2+ab\right)\left(c^2+ac\right)}\ge abc+\sum a\sqrt{bc}\)
Điều này luôn đúng theo BĐT Bunyakovsky:
\(\sum\sqrt{\left(b^2+ab\right)\left(c^2+ac\right)}\ge\sum\left(bc+a\sqrt{bc}\right)=abc+\sum a\sqrt{bc}\)
Dấu = xảy ra khi a=b=c=3
C/Minh đẳng thức:
a) \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{a-1}\) (với a>0, b>0, a≠b)
b)\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) (với a>0, b>0,a≠b)
c) \(\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}=\frac{a+9}{a-9}\) (với a≥0, b≥0,a≠9)