\(a.\frac{{{x^2}}}{{{x^2} + 2yz}} + \frac{{{y^2}}}{{{y^2} + 2zx}} + \frac{{{z^2}}}{{{z^2} + 2xy}} \ge 1\)
\(b.\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} \ge \frac{3}{2}\)
Bài 1: Cho các số thực dương x,y,z. Chứng minh rằng:
\(\frac{x}{\sqrt{2xy+y^2}}+\frac{y}{\sqrt{2yz+z^2}}+\frac{z}{\sqrt{2zx+x^2}}\ge\sqrt{3}\)
Bài 2: Cho a,b,c là các số thực dương thỏa mãn: \(a+b+c=3\)
Tìm min của \(P=\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\)
Bài 3: Cho a,b,c là các số thực dương. Chứng minh rằng:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2+b^2-ab}+\sqrt{b^2+c^2-bc}+\sqrt{c^2+a^2-ca}\)
Rảnh rỗi :D
Cho các số thực dương a,b,c. Tìm Min của:
\(P=\frac{x}{\sqrt{2xy+y^2}}+\frac{y}{\sqrt{2yz+z^2}}+\frac{z}{\sqrt{2zx+x^2}}\)
Cho a,b,c > 0.
Chứng minh:
\(\frac{2xy}{z^2}+\frac{2yz}{x^2}+\frac{2zx}{y^2}-\frac{x}{y}-\frac{y}{x}-\frac{y}{z}-\frac{z}{y}-\frac{z}{x}-\frac{x}{z}\ge0\)
Đặt \(\left\{{}\begin{matrix}\frac{x}{y}=a\\\frac{y}{z}=b\\\frac{z}{x}=c\end{matrix}\right.\) \(\Rightarrow abc=1\)
\(P=\frac{2b}{c}+\frac{2c}{a}+\frac{2a}{b}-a-b-c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\)
\(P=2ab^2+2bc^2+2a^2c-a-b-c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\)
\(ab^2+a\ge2ab\Rightarrow ab^2\ge2ab-a\) ; \(ab^2+\frac{1}{a}\ge2b\Rightarrow ab^2\ge2b-\frac{1}{a}\)
\(\Rightarrow2ab^2\ge2ab+2b-a-\frac{1}{a}\)
Tương tự và cộng lại:
\(\Rightarrow P\ge2\left(ab+ac+bc\right)+2\left(a+b+c\right)-2\left(a+b+c\right)-2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\ge\frac{2\left(ab+ac+bc\right)}{abc}-2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\) hay \(x=y=z\)
Cho x,y,z thỏa mãn \(x^2+y^2+z^2=1\).CMR:
\(\frac{x^2}{1+2yz}+\frac{y^2}{1+2zx}+\frac{z^2}{1+2xy}\ge\frac{3}{5}\)
Ta có: \(\frac{x^2}{1+2yz}+\frac{y^2}{1+2zx}+\frac{z^2}{1+2xy}\)
\(\ge\frac{\left(x+y+z\right)^2}{3+2\left(xy+yz+zx\right)}\ge\frac{\left(x+y+z\right)^2}{3+2\left(x^2+y^2+z^2\right)}\)
\(=\frac{\left(x+y+z\right)^2}{3+2}=\frac{\left(x+y+z\right)^2}{5}\)
Mà \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=3\)
Nên thay vào ngược dấu
=> ch bt lm
Nói chung khá đơn giản. Em chứng minh bất đẳng thức sau đây là được.
\(\frac{x^2}{1+2yz}=\frac{x^2}{x^2+\left(y^2+z^2+2yz\right)}=\frac{x^2}{x^2+\left(y+z\right)^2}\ge\frac{1}{25}\cdot\frac{17x^2-y^2-z^2}{x^2+y^2+z^2}\)
Có thể chứng minnh nó bằng cách: \(f\left(x,y,z\right)=\frac{x^2}{x^2+\left(y+z\right)^2}-\frac{1}{25}\cdot\frac{17x^2-y^2-z^2}{x^2+y^2+z^2}\)
Ta chứng minhL \(f\left(x,y,z\right)\ge f\left(x,\frac{y+z}{2},\frac{y+z}{2}\right)\ge0\) (quy đồng phát là ra nhân tử (y-z)^2 nên hiển nhiên:v)
Tương tự cộng lại. Xong.
Cách Cauchy-SChwarz:
Chứng minh theo trình tự: \(\Sigma\frac{x^2}{x^2+\left(y+z\right)^2}\ge\frac{\left(x^2+y^2+z^2\right)^2}{\Sigma x^2\left[x^2+\left(y+z\right)^2\right]}\ge\frac{3}{5}\)
Mạnh dạn nhân lên xài Cauchy Schwarz thôi ^_^
\(\frac{x^2}{1+2yz}+\frac{y^2}{1+2zx}+\frac{z^2}{1+2xy}\)
\(=\frac{x^4}{x^2+2x^2yz}+\frac{y^4}{y^2+2y^2zx}+\frac{z^4}{1+2z^2xy}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+2xyz\left(x+y+z\right)}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+\frac{2\left(xy+yz+zx\right)^2}{3}}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+\frac{2\left(x^2+y^2+z^2\right)^2}{3}}=\frac{3}{5}\)
Đẳng thức xảy ra tại x=y=z=\(\frac{1}{\sqrt{3}}\)
Cho \(a,b,c\ge0\)
CM: \(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2zx}+\frac{z^2}{z^2+2xy}\ge1\)
\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2zx}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2zx+z^2+2xy}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Cho x,y,z dương, x+y+z=1. Chứng minh:
\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}\ge9\)
Bạn tự c/m BĐT : \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Dấu " = " xảy ra ta có:
\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1\right)^2}{x^2+y^2+2yz+2zx}+\frac{1}{z^2+2xy}\)\(\ge\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}=\frac{9}{1}=9\)
Bạn tự giải dấu bằng nhé.
a, Cho các số a,b,c,d nguyên dương đôi một khác nhau thoả mãn:
\(\frac{2a+b}{a+b}+\frac{2b+c}{b+c}+\frac{2c+d}{c+d}+\frac{2d+a}{d+a}=6\). CMR: A = abcd là số chính phương
b, Giải phương trình: \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
c, Cho x,y,z dương và x + y + z = 1. CMR: \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge9\)
d, Tìm nghiệm nguyên dương của phương trình: \(\frac{2016}{x+y}+\frac{x}{y+2015}+\frac{y}{4031}+\frac{2015}{x+2016}=2\)
a) Ta có:
\(\frac{2a+b}{a+b}+\frac{2b+c}{b+c}+\frac{2c+d}{c+d}+\frac{2d+a}{d+a}=6\)
\(\Leftrightarrow\left[\left(\frac{2a+b}{a+b}-1\right)+\left(\frac{2b+c}{b+c}-1\right)-1\right]+\left[\left(\frac{2c+d}{c+d}-1\right)+\left(\frac{2d+a}{d+a}-1\right)-1\right]=0\)
\(\Leftrightarrow\left(\frac{a}{a+b}+\frac{b}{b+c}-1\right)+\left(\frac{c}{c+d}+\frac{d}{d+a}-1\right)=0\)
\(\Leftrightarrow\left(\frac{a.\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}+\frac{b.\left(a+b\right)}{\left(a+b\right).\left(b+c\right)}-\frac{\left(a+b\right).\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{c.\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}+\frac{d.\left(c+d\right)}{\left(c+d\right).\left(d+a\right)}-\frac{\left(c+d\right).\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\left(\frac{ab+ac}{\left(a+b\right).\left(b+c\right)}+\frac{ab+b^2}{\left(a+b\right).\left(b+c\right)}-\frac{ab+ac+b^2+bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac}{\left(c+d\right).\left(d+a\right)}+\frac{cd+d^2}{\left(c+d\right).\left(d+a\right)}-\frac{cd+ac+d^2+ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\left(\frac{ab+ac+ab+b^2-ab-ac-b^2-bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac+cd+d^2-cd-ac-d^2-ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}+\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}=0\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=-\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=\frac{ad-cd}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{b.\left(a-c\right)}{\left(a+b\right).\left(b+c\right)}=\frac{d.\left(a-c\right)}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{b}{\left(a+b\right).\left(b+c\right)}=\frac{d}{\left(c+d\right).\left(d+a\right)}\) (vì \(a;b;c;d\) là số nguyên dương).
\(\Leftrightarrow b\left(c+d\right).\left(d+a\right)=d\left(a+b\right).\left(b+c\right)\)
\(\Leftrightarrow\left(bc+bd\right).\left(d+a\right)=\left(ad+bd\right).\left(b+c\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow bd^2+abc=b^2d+acd\)
\(\Leftrightarrow bd^2-b^2d=acd-abc\)
\(\Leftrightarrow bd.\left(d-b\right)=ac.\left(d-b\right)\)
\(\Leftrightarrow bd.\left(d-b\right)-ac.\left(d-b\right)=0\)
\(\Leftrightarrow\left(d-b\right).\left(bd-ac\right)=0\)
Vì \(a;b;c;d\) là số nguyên dương.
\(\Rightarrow d-b>0\)
\(\Rightarrow d-b\ne0.\)
\(\Leftrightarrow bd-ac=0\)
\(\Leftrightarrow bd=ac.\)
Lại có:
\(A=abcd\)
\(\Rightarrow A=ac.bd\)
\(\Rightarrow A=ac.ac\)
\(\Rightarrow A=\left(ac\right)^2.\)
\(\Rightarrow A=abcd\) là số chính phương (đpcm).
Chúc bạn học tốt!
1.Cho a,b,c dương, a+b+c≤1.CMR: \(\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\ge10\)
2.Cho a,b, c >0. CMR: \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge\sqrt{82};x+y+z\le1\)
2.
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1+9^2\right)\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow82\cdot\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow\sqrt{82}\cdot\sqrt{x^2+\frac{1}{x^2}}\ge x+\frac{9}{x}\)
Tương tự ta cũng có :
\(\sqrt{82}\cdot\sqrt{y^2+\frac{1}{y^2}}\ge y+\frac{9}{y}\)
\(\sqrt{82}\cdot\sqrt{z^2+\frac{1}{z^2}}\ge z+\frac{9}{z}\)
Cộng theo vế của các bất đẳng thức ta được :
\(\sqrt{82}\cdot\left(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\right)\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\)
\(\Leftrightarrow\sqrt{82}\cdot P\ge x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}\)(1)
Mặt khác áp dụng bất đẳng thức Cauchy ta có :
\(x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}=81x+\frac{9}{x}+81y+\frac{9}{y}+81z+\frac{9}{z}-80x-80y-80z\)
\(\ge2\sqrt{\frac{81x\cdot9}{x}}+2\sqrt{\frac{81y\cdot9}{y}}+2\sqrt{\frac{81z\cdot9}{z}}-80\left(x+y+z\right)\)
\(\ge2\sqrt{729}+2\sqrt{729}+2\sqrt{729}-80\cdot1\)
\(=82\) (2)
Từ (1) và (2) suy ra \(\sqrt{82}\cdot P\ge82\)
\(\Leftrightarrow P\ge\sqrt{82}\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
1.
Áp dụng bất đẳng thức Cauchy :
\(\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\)
\(=a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\)
\(=9a+\frac{1}{a}+9b+\frac{1}{b}+9c+\frac{1}{c}-8a-8b-8c\)
\(\ge2\sqrt{\frac{9a}{a}}+2\sqrt{\frac{9b}{b}}+2\sqrt{\frac{9c}{c}}-8\left(a+b+c\right)\)
\(\ge3\cdot2\sqrt{9}-8=10\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Cho x,y,z là những số thực dương và các số thực a,b,c
Chứng minh: \(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)\left(x+y+z\right)\ge\left(a+b+c\right)^2\)
Áp dụng bất đẳn thức Cauchy-Schwarz ta có:
\(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)\left(x+y+z\right)=\)\(\left[\frac{a^2}{\left(\sqrt{x}\right)^2}+\frac{b^2}{\left(\sqrt{y}\right)^2}+\frac{c^2}{\left(\sqrt{z}\right)^2}\right]\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)
\(\ge\left(\frac{a}{\sqrt{x}}.\sqrt{x}+\frac{b}{\sqrt{y}}.\sqrt{y}+\frac{c}{\sqrt{z}}.\sqrt{z}\right)=\left(a+b+c\right)\)\(\left(đpcm\right)\)
ấy chết em quên ko có mũ 2
Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)\left(x+y+z\right)\)\(=\left[\frac{a^2}{\left(\sqrt{x}\right)^2}+\frac{b^2}{\left(\sqrt{y}\right)^2}+\frac{c^2}{\left(\sqrt{z}\right)^2}\right]\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)
\(\ge\left(\frac{a}{\sqrt{x}}.\sqrt{x}+\frac{b}{\sqrt{y}}.\sqrt{y}+\frac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\left(đpcm\right)\)