\(\dfrac{1}{1-x}\ge2\)
\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2x-2\right)\cdot2x}=\dfrac{3}{16}\left(x\in Z,x\ge2\right)\)
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+......+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{3}{16}\)
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{3}{16}\)
\(\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{3}{16}:\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{3}{8}\)
\(\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{3}{8}\)
\(\dfrac{1}{2x}=\dfrac{1}{8}\)
⇒x=8:2=4
cho x,y,z>0 và x3+y3+z3=1.
CMR:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\)
Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)
CM CÁC BẤT ĐẲNG THỨC SAU
A) \(X+\dfrac{1}{X}\ge2\) (X>0)
B) \(\dfrac{A}{B}+\dfrac{B}{A}\ge2\) (AB>0)
CM CÁC BẤT ĐẲNG THỨC SAU
A) \(X+\dfrac{1}{X}\ge2\) (X>0)
B) \(\dfrac{A}{B}+\dfrac{B}{A}\ge2\) (AB>0)
Bạn hỏi câu này có lẽ bạn chưa biết BĐT côsi, mk sẽ trình bày từ bước chứng minh BĐT
Ta có: \(\left(m-n\right)^2\ge0\)
<=> \(m^2-2m.n+n^2\ge0\)
<=> \(m^2+2m.n+n^2-4m.n\ge0\)
<=> \(\left(m+n\right)^2\ge4m.n\)
=> \(m+n\ge2\sqrt{m.n}\) ( BĐT côsi)
a, Áp dụng BĐT côsi ta có:
\(\dfrac{1}{x}+x\ge2\sqrt{\dfrac{1}{x}.x}=2\)
vậy \(\dfrac{1}{x}+x\ge2\) (x>0)
b, Áp dụng BĐT côsi ta có:
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
vậy \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) với a, b >0
-----------Chúc bạn học tốt -------------
Cho x,y,z > -1. Chứng minh A = \(\dfrac{1+x^2}{1+y+z^2}+\dfrac{1+y^2}{1+z+x^2}+\dfrac{1+z^2}{1+x+y^2}\ge2\)
\(\sum\dfrac{x^2+1}{\left(z^2+1\right)+y}\ge\sum\dfrac{x^2+1}{\left(z^2+1\right)+\dfrac{y^2+1}{2}}\)
Áp dụng BĐT AM-GM ta có:
\(y\le\dfrac{y^2+1}{2}\Rightarrow\dfrac{1+x^2}{1+y+z^2}\ge\dfrac{1+x^2}{1+\dfrac{y^2+1}{2}+z^2}\)
Tương tự cho 2 BĐT còn lại thì viết lại dc thành
\(\dfrac{1+x^2}{z^2+1+\dfrac{y^2+1}{2}}+\dfrac{1+y^2}{x^2+1+\dfrac{z^2+1}{2}}+\dfrac{1+z^2}{y^2+1+\dfrac{x^2+1}{2}}\)
Đặt \(\left\{{}\begin{matrix}x^2+1=a\\y^2+1=b\\z^2+1=c\end{matrix}\right.\)\(\left(a,b,c>0\right)\) thì ta có:
\(\dfrac{a}{c+\dfrac{b}{2}}+\dfrac{b}{a+\dfrac{c}{2}}+\dfrac{c}{b+\dfrac{a}{2}}\ge2\)
\(\Leftrightarrow\dfrac{a}{2c+b}+\dfrac{b}{2a+c}+\dfrac{c}{2b+a}\ge1\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{2ac+ab}+\dfrac{b^2}{2ab+bc}+\dfrac{c^2}{2bc+ca}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+2ab+2bc+2ca}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1=VP\)
Tìm x biết :
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right)2x}=\dfrac{1}{8}\left(x\in N,x\ge2\right)\)
mình ko biết mình làm đúng hay sai bạn nhé, mong mọi người góp ý
= 1/2.( 1/2.4+1/4.6+....+1/(2x-2)2x)=1/8
= 1/2.(1/2-1/4+1/4-1/6+....+1/(2x-2)-1/2x)=1/8
= 1/2.( 1/2-1/2x)=1/8
( 1/2-1/2x)=1/8:1/2
1/2-1/2x=1/4
1/2x =1/2-1/4
1/2x =1/4
2x = 4
x =4:2
x =2
Cho ba số dương x;y;z thỏa mãn:\(\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}\ge2\).Tìm GTLN của P=x+y.
Chứng minh rằng với mọi x, y, z > 0 ta có: \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\ge2+\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Ta có:
\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{y}{z}+\dfrac{x}{z}+\dfrac{z}{x}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Ta có:
\(\dfrac{x}{y}+\dfrac{x}{y}+1\ge3\sqrt[3]{\dfrac{x^2}{y^2}}\)
Tương tự ...
Cộng lại ta có:
\(2\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\ge3\left(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\right)\)
\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\)
Do đó ta chỉ cần chứng minh:
\(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow\left(\sqrt[3]{\dfrac{x}{y}}-\sqrt[3]{\dfrac{x}{z}}\right)^2+\left(\sqrt[3]{\dfrac{y}{x}}-\sqrt[3]{\dfrac{y}{z}}\right)^2+\left(\sqrt[3]{\dfrac{z}{x}}-\sqrt[3]{\dfrac{z}{y}}\right)^2\ge0\) (luôn đúng)
Cho x,y,z>0 và \(\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}\ge2\)
Chứng minh: xyz≤\(\dfrac{1}{8}\)
Ta có \(\dfrac{1}{1+x}\ge1-\dfrac{1}{1+y}+1-\dfrac{1}{1+x}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\)
\(\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\)
Chứng minh tương tự, ta có
\(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(z+1\right)\left(x+1\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)
Nhân cả 3 cua 3 BĐT cùng chiều, ta có
\(\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\dfrac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Rightarrow xuz\le\dfrac{1}{8}\left(ĐPCM\right)\)