Cho a,b,c là các số thực dương
Chứng minh
Cho a,b,c là các số thực dương
Chứng minh
BĐT bên trái: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\sqrt{\dfrac{2a}{b+c}}+\sqrt{\dfrac{2b}{c+a}}+\sqrt{\dfrac{2c}{a+b}}\)
Ta có: \(\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\left(ab+bc+ca\right)\ge\left(a+b+c\right)^2\)
\(\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
Nhân vế với vế và rút gọn:
\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Lại có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\dfrac{2}{b+c}+\dfrac{2}{c+a}+\dfrac{2}{a+b}\right)\ge\left(\sqrt{\dfrac{2a}{b+c}}+\sqrt{\dfrac{2b}{c+a}}+\sqrt{\dfrac{2c}{a+b}}\right)^2\)
\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\sqrt{\dfrac{2a}{b+c}}+\sqrt{\dfrac{2b}{c+a}}+\sqrt{\dfrac{2c}{a+b}}\) (đpcm)
BĐT bên phải:
\(\sqrt{\dfrac{2a}{b+c}}+\sqrt{\dfrac{2b}{c+a}}+\sqrt{\dfrac{2c}{a+b}}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2}\)
Ta có:
\(VT=\dfrac{2a}{\sqrt{2a.\left(b+c\right)}}+\dfrac{2b}{\sqrt{2b\left(c+a\right)}}+\dfrac{2c}{\sqrt{2c\left(a+b\right)}}\)
\(\ge\dfrac{4a}{2a+b+c}+\dfrac{4b}{2b+c+a}+\dfrac{4c}{2c+a+b}\)
\(=\dfrac{4a^2}{2a^2+ab+ac}+\dfrac{4b^2}{2b^2+bc+ab}+\dfrac{4c^2}{2c^2+ac+bc}\)
\(\ge\dfrac{4\left(a+b+c\right)^2}{2a^2+2b^2+2c^2+2\left(ab+bc+ca\right)}\ge\dfrac{4\left(a+b+c\right)^2}{2a^2+2b^2+2c^2+2\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2}\) (đpcm)
\(2+2cosx=2\left(1+cosx\right)=2\left(1+2cos^2\dfrac{x}{2}-1\right)=4cos^2\dfrac{x}{2}\)
\(\Rightarrow\sqrt{2+2cosx}=2cos\dfrac{x}{2}=2cos\dfrac{x}{2^1}\) (1 dấu căn)
\(\Rightarrow\sqrt{2+\sqrt{2+2cosx}}=\sqrt{2+2cos\dfrac{x}{2}}=2cos\dfrac{x}{4}=2cos\dfrac{x}{2^2}\) (2 dấu căn)
Từ đó ta có: \(P=2cos\dfrac{x}{2^{2020}}\)
ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)
\(\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14x-5\right)< 0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)< 0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)< 0\)
Do \(\dfrac{3}{\sqrt{3x+1}+1}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\in\left[-\dfrac{1}{3};6\right]\) nên BPT tương đương:
\(x-5< 0\Leftrightarrow x< 5\)
Vậy tập nghiệm của BPT là: \(-\dfrac{1}{3}\le x< 5\)
cho bất phương trình \(f\left(x\right)\le g\left(x\right),x_0\) là một nghiệm của bất phương trình \(f\left(x\right)\le g\left(x\right)\) nếu:
A. f(x0)=g(x0) đúng
B. f(x0) >= g(xo) đúng
C. f(x0) <= g(x0) sai
D. f(x0) > g(x0) đúng
Cho 2 số thực dương thỏa mãn x+y+3xy=1
Tìm GTLN của biểu thức A= \(\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{3xy}{x+y}\)
\(1=x+y+3xy\le x+y+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow3\left(x+y\right)^2+4\left(x+y\right)-4\ge0\)
\(\Rightarrow3\left(x+y+2\right)\left(x+y-\dfrac{2}{3}\right)\ge0\)
\(\Rightarrow x+y\ge\dfrac{2}{3}\) \(\Rightarrow\dfrac{1}{x+y}\le\dfrac{3}{2}\)
Đồng thời: \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^2=\dfrac{2}{9}\)
\(\Rightarrow-\left(x^2+y^2\right)\le-\dfrac{2}{9}\)
Từ đó ta có:
\(A=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1-\left(x+y\right)}{x+y}=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1}{x+y}-1\)
\(A\le\sqrt{2\left[2-\left(x^2+y^2\right)\right]}+\dfrac{1}{x+y}-1\le\sqrt{2\left(2-\dfrac{2}{9}\right)}+\dfrac{3}{2}-1=\dfrac{3+8\sqrt{2}}{6}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)
Cho a,b,c là các sô thực dương thỏa mãn a+b+c=1
CMR
Với a;b;c dương:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-\sqrt[3]{abc}.\sqrt[3]{ab.bc.ca}\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\dfrac{1}{3}\left(a+b+c\right).\dfrac{1}{3}\left(ab+bc+ca\right)\)
\(=\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
Đặt vế trái BĐT là P, ta có:
\(\dfrac{ab}{1-c^2}=\dfrac{ab}{\left(1-c\right)\left(1+c\right)}=\dfrac{ab}{\left(a+b\right)\left(a+c+b+c\right)}=\dfrac{ab}{\sqrt{a+b}.\sqrt{a+b}\left(a+c+b+c\right)}\)
\(\le\dfrac{ab}{\sqrt[]{2\sqrt[]{ab}}.\sqrt[]{a+b}.2\sqrt[]{\left(a+c\right)\left(b+c\right)}}=\dfrac{\sqrt[4]{\left(ab\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
Tương tự:
\(\dfrac{bc}{1-a^2}\le\dfrac{\sqrt[4]{\left(bc\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
\(\dfrac{ca}{1-b^2}\le\dfrac{\sqrt[4]{\left(ca\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
Cộng vế:
\(P\le\dfrac{\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
Nên ta chỉ cần chứng minh:
\(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\le\dfrac{3}{2\sqrt[]{2}}\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow\left(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\right)^2\le\dfrac{9}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Mà \(\dfrac{9}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)\)
Nên ta chỉ cần chứng minh:
\(\left(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\right)^2\le\left(a+b+c\right)\left(ab+bc+ca\right)\)
Thật vậy:
\(\left(\sqrt[4]{ab}.\sqrt[]{ab}+\sqrt[4]{bc}.\sqrt[]{bc}+\sqrt[4]{ca}.\sqrt[]{ca}\right)^2\le\left(\sqrt[]{ab}+\sqrt[]{bc}+\sqrt[]{ca}\right)\left(ab+bc+ca\right)\)
\(\le\left(a+b+c\right)\left(ab+bc+ca\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho x,y,z là các số thực dương
Ta có:
\(VT=\sqrt{x+z}\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}}+\sqrt{x+y}\sqrt{\dfrac{y}{\left(x+y\right)\left(y+z\right)}}+\sqrt{y+z}\sqrt{\dfrac{z}{\left(x+z\right)\left(y+z\right)}}\)
\(\Rightarrow VT^2\le\left(x+z+x+y+y+z\right)\left(\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(x+y\right)\left(y+z\right)}+\dfrac{z}{\left(x+z\right)\left(y+z\right)}\right)\)
\(\Rightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác ta có:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(\Rightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+zx\right)}{\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)}=\dfrac{9}{2}\)
\(\Rightarrow VT\le\dfrac{3\sqrt{2}}{2}\) (đpcm)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z\)
Giúp mình với
1: Δ=(2m-2)^2-4(m+5)
=4m^2-8m+4-4m-20
=4m^2-12m-16
Để f(x)>0 với mọi x thì 4m^2-12m-16<0 và 1>0
=>m^2-3m-4<0
=>(m-4)(m+1)<0
=>-1<m<4
giúp mình vs
1A
2A
3C
4D
5: A,D đều đúng nha bạn
7A
8B