Cho a,b,c >0 và \(a+b+c\le1\)
Chứng minh rằng : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+2ab}\ge9\)
Cho a, b, c > 0 và a + b + c \(\le\)1. Chứng minh rằng:
\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)
Cho a,b,c lớn hơn 0 và\(a+b+c\le1\)
CM; \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)
Theo bất đẳng thức Cauchy-Schwartz ta có
\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+2bc+b^2+2ca+c^2+2ab}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9.\)
cho \(a,b,c\ge0;a+b+c\le1\).CMR: \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge9\)
cho a,b,c>0 và \(a+b+c\le1\)
cmr \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\ge9\)
Với \(a+b+c\le1\) và a, b, c >0
CMR:\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ba}\ge9\)
ĐK:\(a+b+c\le1|a,b,c>0\)
Chỉ có TH \(a=b=c=\frac{1}{3}\)\(\Rightarrow TH:a+b+c=1\)
\(\Rightarrow\frac{1}{\left(\frac{1}{3}\right)^2+2.\frac{1}{3}.\frac{1}{3}}+\frac{1}{\left(\frac{1}{3}\right)^2+2.\frac{1}{3}.\frac{1}{3}}+\frac{1}{\left(\frac{1}{3}\right)^2+2.\frac{1}{3}.\frac{1}{3}}\ge9\)\(=\frac{1}{\left(\frac{1}{3}\right)^2+2\left(\frac{1}{3}\right)^2}3\ge9\)\(=\frac{1}{\left(\frac{1}{3}\right)^2\left(2+1\right)}3\ge9\)\(=\frac{1}{\left(\frac{1}{3}\right)^2.3}3\ge9\)\(=\frac{1}{\frac{1}{3}.\frac{1}{3}.3}3\ge9\)\(=\frac{1}{\frac{1}{3}}3\ge9\)\(=\frac{3}{\frac{1}{3}}\ge9\)\(=3:1:3\ge9\)\(=1\ge9\)( loại )
Vậy không thể CMR \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ba}\ge9\).
Áp dụng bất đẳng thức Cauchy - Schwarz ta có:
\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)
\(=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1^2}=9\)
Dấu "=" xảy ra khi: a = b = c = 1/3
Cho a,b,c là các số dương và \(a+b+c\le1\)
Chứng minh \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\)9
\(\frac{1}{a^2+2ab}+\frac{1}{b^2+2bc}+\frac{1}{c^2+2ab}\ge\frac{9}{a^2+2ab+b^2+2bc+c^2+2ab}=\frac{9}{\left(a+b+c\right)^2}\ge9\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Cho \(a,b,c>0\) và \(a+b+c=1\) CMR \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)
Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) với \(x=a^2+2bc;y=b^2+2ac;z=c^2+2ab\)
Ta có : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=\frac{9}{\left(a+b+c\right)^2}\)
\(\Rightarrow\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)( Vì a + b + c = 1)
cho a, b, c là các số không âm. Chứng minh rằng:
\(\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\)
Ta chứng minh bất đẳng thức: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) (a,b,c,x,y,z dương) (Hệ quả của bất đẳng thức Cauchy-Schwarz (Bunyakovsky))
\(\left[\frac{a^2}{\left(\sqrt{x}\right)^2}+\frac{b^2}{\left(\sqrt{y}\right)^2}+\frac{c^2}{\left(\sqrt{z}\right)^2}\right]\left[\left(\sqrt{x}\right)^2+\sqrt{y}^2+\sqrt{z^2}\right]\ge a^2+b^2+c^2\)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Ta có:
\(A=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(2A=\frac{2bc}{a^2+2bc}+\frac{2ca}{b^2+2ac}+\frac{2ab}{c^2+2ab}\)
\(=\frac{a^2+2bc-a^2}{a^2+2bc}+\frac{b^2+2ca-b^2}{b^2+2ac}+\frac{c^2+2ab-c^2}{c^2+2ab}\)
\(=3-\left(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\right)\)
\(\le3-\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}=3-\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=3-1=2\)
=> A<=1
a,b,c dương
Ta viết lại BĐT thành: \(\frac{1}{\frac{a^2}{bc}+2}+\frac{1}{\frac{b^2}{ca}+2}+\frac{1}{\frac{c^2}{ab}+2}\le1\)
Đặt \(\frac{a^2}{bc}=x;\frac{b^2}{ca}=y;\frac{c^2}{ab}=z\Rightarrow\hept{\begin{cases}x,y,z>0\\xyz=1\end{cases}}\)và ta cần chứng minh \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\le1\)
Xét biểu thức\(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\) \(\frac{\left(y+2\right)\left(z+2\right)+\left(z+2\right)\left(x+2\right)+\left(x+2\right)\left(y+2\right)}{\left(x+2\right)\left(y+2\right)\left(z+2\right)}\)
\(=\frac{\left(yz+2y+2z+4\right)+\left(zx+2z+2x+4\right)+\left(xy+2x+2y+4\right)}{\left(xy+2x+2y+4\right)\left(z+2\right)}\)
\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+2\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+\left(xy+yz+zx\right)+\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(\le\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+3\sqrt{\left(xyz\right)^2}+\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}=1\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c
Ta viết lại BĐT thành: \(\frac{1}{\frac{a^2}{bc}+2}+\frac{1}{\frac{b^2}{ca}+2}+\frac{1}{\frac{c^2}{ab}+2}\le1\)
Đặt \(\frac{a^2}{bc}=x^2;\frac{b^2}{ca}=y^2;\frac{c^2}{ab}=z^2\)thì \(xyz=1\)
Khi đó BĐT chuyển thành dạng:\(\frac{1}{x^2+2}+\frac{1}{y^2+2}+\frac{1}{z^2+2}\le1\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x^2+2}+\frac{1}{2}-\frac{1}{y^2+2}+\frac{1}{2}-\frac{1}{z^2+2}\ge\frac{3}{2}-1=\frac{1}{2}\)
\(\Leftrightarrow\frac{x^2}{x^2+2}+\frac{y^2}{y^2+2}+\frac{z^2}{z^2+2}\ge1\)
Áp dụng BĐT Bunyakovsky dạng phân thức, ta được: \(\frac{x^2}{x^2+2}+\frac{y^2}{y^2+2}+\frac{z^2}{z^2+2}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+6}\)
Đến đây, ta cần chỉ ra rằng \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+6}\ge1\Leftrightarrow xy+yz+zx\ge3\)(Đúng theo BĐT AM - GM vì \(xy+yz+zx\ge3\sqrt[3]{\left(xyz\right)^2}=3\)
Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c
Cho a,b,c khác nhau đôi một và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Rút gọn
a) \(A=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
b) \(B=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
c) \(C=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)