Bài 1: Căn bậc hai
Các câu hỏi tương tự
cho a,b,c>0
CMR:
1) \(a+b+\dfrac{1}{4}\ge\sqrt{a+b}\)
2) \(\left(a+b+\dfrac{1}{2}\right)^2+\left(b+c+\dfrac{1}{2}\right)^2+\left(c+a+\dfrac{1}{2}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
c/m bất đảng thức :
a)dfrac{a}{3b}+dfrac{bleft(a+bright)}{a^2+ab+b^2}
b)dfrac{a}{b^2}+dfrac{b}{a^2}+dfrac{16}{a+b}ge5left(dfrac{1}{a}+dfrac{1}{b}right)
c)dfrac{a}{2b}+dfrac{2b}{a+b}+dfrac{ab^2}{2left(a^3+2b^3right)}gedfrac{5}{3}
d)dfrac{a}{4b^2}+dfrac{2b}{left(a+bright)^2}gedfrac{9}{4left(a+2bright)}
e)dfrac{2}{a^2+ab+b^2}+dfrac{1}{3b^2}gedfrac{9}{left(a+2bright)^2}
Đọc tiếp
c/m bất đảng thức :
a)\(\dfrac{a}{3b}+\dfrac{b\left(a+b\right)}{a^2+ab+b^2}\)
b)\(\dfrac{a}{b^2}+\dfrac{b}{a^2}+\dfrac{16}{a+b}\ge5\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
c)\(\dfrac{a}{2b}+\dfrac{2b}{a+b}\)+\(\dfrac{ab^2}{2\left(a^3+2b^3\right)}\ge\dfrac{5}{3}\)
d)\(\dfrac{a}{4b^2}+\dfrac{2b}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+2b\right)}\)
e)\(\dfrac{2}{a^2+ab+b^2}+\dfrac{1}{3b^2}\ge\dfrac{9}{\left(a+2b\right)^2}\)
ta có sqrt{left(1+a^3right)left(1+b^3right)}sqrt{left(1+aright)left(a^2-a+1right)}.sqrt{left(1+bright)left(b^2-b+1right)}
Mà sqrt{left(a+1right)left(a^2-a+1right)}ledfrac{a+1+a^2-a+2}{2}dfrac{a^2+2}{2}
Tương tự thì sqrt{left(1+a^3right)left(1+b^3right)}ledfrac{left(a^2+2right)left(b^2+2right)}{4}Rightarrowdfrac{a^2}{sqrt{left(1+a^3right)left(1+B^3right)}}gedfrac{4a^2}{left(a^2+2right)left(b^2+2right)}...
Đọc tiếp
ta có \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}=\sqrt{\left(1+a\right)\left(a^2-a+1\right)}.\sqrt{\left(1+b\right)\left(b^2-b+1\right)}\)
Mà \(\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\le\dfrac{a+1+a^2-a+2}{2}=\dfrac{a^2+2}{2}\)
Tương tự thì \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}\le\dfrac{\left(a^2+2\right)\left(b^2+2\right)}{4}\Rightarrow\dfrac{a^2}{\sqrt{\left(1+a^3\right)\left(1+B^3\right)}}\ge\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\)
=\(\dfrac{4a^2\left(c^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
Tương tự rồi + vào, ta có
...\(\ge4\dfrac{a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
ta cần chứng minh \(3\left[a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)\right]\ge\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\)
đến đây nhân tung ra và dùng cô-si tiếp
cho a,b,c>0. chứng minh rằng
\(\dfrac{1}{a\left(b+1\right)}+\dfrac{1}{b\left(c+1\right)}+\dfrac{1}{c\left(a+1\right)}\ge\dfrac{3}{abc+1}\)
Câu 1 ) Cho a,b,cin R . Chứng minh rằng :
Mleft(a^2+1right)left(b^2+1right)left(c^2+1right)gedfrac{3left(a+b+cright)^2}{4}
Câu 2 ) Cho a0;b0;a+ble1 . Tìm GTNN của biểu thức :
A dfrac{2}{a^2+b^2}+dfrac{35}{ab}+2ab
Câu 3) Cho a0;b0 . Chứng minh rằng : left(4a^2+b^2right)left(dfrac{1}{a^2}+dfrac{1}{4b^2}right)ge4
Đọc tiếp
Câu 1 ) Cho \(a,b,c\in R\) . Chứng minh rằng :
M=\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{3\left(a+b+c\right)^2}{4}\)
Câu 2 ) Cho \(a>0;b>0;a+b\le1\) . Tìm GTNN của biểu thức :
A = \(\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)
Câu 3) Cho \(a>0;b>0\) . Chứng minh rằng : \(\left(4a^2+b^2\right)\left(\dfrac{1}{a^2}+\dfrac{1}{4b^2}\right)\ge4\)
CMR:
A=\(\dfrac{1}{3\left(1+\sqrt{2}\right)}+\dfrac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\dfrac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)<\(\dfrac{1}{2}\)
Cho x,y,z>0. CM: \(\dfrac{xy}{z^2\left(x+y\right)}+\dfrac{yz}{x^2\left(y+z\right)}+\dfrac{zx}{y^2\left(z+x\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
1) gpt \(x^2+3x\sqrt{\dfrac{x^2+1}{x}}=10x-1\)
2) ghpt \(\left\{{}\begin{matrix}x^2+y^2+2\left(x+y\right)=6\\xy\left(x+2\right)\left(y+2\right)=9\end{matrix}\right.\)
3) cho a,b,c dương thỏa abc=1
CMR \(\dfrac{2}{a^2\left(b+c\right)}+\dfrac{2}{b^2\left(c+a\right)}+\dfrac{2}{c^2\left(a+b\right)}\ge3\)
Cho các số thực dương x, y, z thỏa mãn : xyz=1.CMR:
\(\dfrac{1}{\left(\sqrt{xy}+\sqrt{x}+1\right)^2}+\dfrac{1}{\left(\sqrt{yz}+\sqrt{y}+1\right)^2}+\dfrac{1}{\left(\sqrt{xz}+\sqrt{z}+1\right)^2}\ge\dfrac{1}{3}\)
Giúp mk với , mk sắp thi r...