\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}\sqrt{c^2+a^2}=\sqrt{2011}cmr\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2011}{2}}\)
Cho a,b,c>0 t/m: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\)
CMR:\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2011}{2}}\)
Áp dụng bất đẳng thức bunyakovsky: \(\left(b+c\right)^2\le2\left(b^2+c^2\right)\Leftrightarrow b+c\le\sqrt{2\left(b^2+c^2\right)}\)
tương tự với các cặp còn lại , ta thu được \(VT\ge\frac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\frac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\frac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
Đặt \(\hept{\begin{cases}\sqrt{b^2+c^2}=x\\\sqrt{a^2+c^2}=y\\\sqrt{a^2+b^2}=z\end{cases}}\)(\(x,y,z\ge0\)và \(x+y+z=\sqrt{2011}\))\(\Leftrightarrow\hept{\begin{cases}a^2=\frac{y^2+z^2-x^2}{2}\\b^2=\frac{x^2+z^2-y^2}{2}\\c^2=\frac{x^2+y^2-z^2}{2}\end{cases}}\)
\(VT\ge\frac{y^2+z^2-x^2}{2\sqrt{2}x}+\frac{x^2+z^2-y^2}{2\sqrt{2}y}+\frac{x^2+y^2-z^2}{2\sqrt{2}z}\)
\(=\frac{1}{2\sqrt{2}}\left(\frac{y^2+z^2-x^2}{x}+\frac{z^2+x^2-y^2}{y}+\frac{x^2+y^2-z^2}{z}\right)=\frac{1}{2\sqrt{2}}\left(\frac{y^2}{x}+\frac{z^2}{x}+\frac{z^2}{y}+\frac{x^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-x-y-z\right)\)
ÁP dụng bất đẳng thức cauchy-schwarz:
\(\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{y^2}{z}+\frac{x^2}{x}\ge\frac{\left(2x+2y+2z\right)^2}{2x+2y+2z}=2x+2y+2z\)
do đó \(VT\ge\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2}\sqrt{\frac{2011}{2}}\)( vì \(x+y+z=\sqrt{2011}\))
đẳng thức xảy ra khi \(x=y=z=\frac{\sqrt{2011}}{3}\)hay \(a=b=c=\frac{1}{3}\sqrt{\frac{2011}{2}}\)
cho ba số duong a,b,c tỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2=\sqrt{2011}}\)chứng minh rằng \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2011}{2}}\)
với a,b,c dương thỏa
\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2015}\\CMR:\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2015}{2}}\)
Thắng Nguyễn Phần cuối cùng viết rõ ra một chút :
\(2\sqrt{2}\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)\ge\frac{y^2+z^2-x^2}{x}+\frac{y^2+x^2-z^2}{z}+\frac{x^2+z^2-y^2}{y}\)
\(\frac{y^2}{x}+\frac{z^2}{x}+\frac{y^2}{z}+\frac{x^2}{z}+\frac{x^2}{y}+\frac{z^2}{y}-\sqrt{2015}\ge\frac{\left[2\left(x+y+z\right)\right]^2}{2\left(x+y+z\right)}-\sqrt{2015}=\sqrt{2015}\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\sqrt{2015}}{2\sqrt{2}}=\frac{1}{2}\sqrt{\frac{2015}{2}}\)
Đặt \(\sqrt{a^2+b^2=z};\sqrt{a^2+c^2}=y;\sqrt{b^2+c^2}=x\left(x;y;z>0\right)\)
\(\Rightarrow a^2=\frac{y^2+z^2-x^2}{2};b=\frac{x^2+z^2-y^2}{2};c=\frac{x^2+y^2-z^2}{2}\)
Theo đề \(x+y+z=\sqrt{2015}\)
Ta có:\(b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}\cdot x\)\(\Rightarrow\frac{a^2}{b+c}\ge\frac{y^2+z^2-x^2}{2\sqrt{2}\cdot x}\)
Tương tự cho 2 cái còn lại rồi, cộng lại:
\(VT\cdot2\sqrt{2}\ge\sqrt{2015}\Rightarrow VT\ge\frac{1}{2}\sqrt{\frac{2015}{2}}\)
Cho a,b,c là các số dương thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2019}\)
CMR: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\sqrt{\frac{2019}{8}}\)
\(VT\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2019}\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\) \(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)
\(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\)
\(2\sqrt{2}VT\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)
\(2\sqrt{2}VT\ge\dfrac{4\left(x+y+z\right)^2}{2x+2y+2z}-\left(x+y+z\right)=x+y+z=\sqrt{2019}\)
\(\Rightarrow VT\ge\dfrac{\sqrt{2019}}{2\sqrt{2}}=\sqrt{\dfrac{2019}{8}}\) (đpcm)
Cho a,b,c>0 tm: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+ \sqrt{a^2+c^2}=\sqrt{2018}\)
CMR \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \ge\frac{1}{2}\sqrt{2009}\)
Tuogw tựCâu hỏi của Nue nguyen - Toán lớp 10 | Học trực tuyến
Cho a,b,c > 0 thỏa mãn \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{1}{2}\). CMR:
\(\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)
\(P=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\)
\(\Rightarrow P\ge\sqrt{\frac{\left(x+2y\right)^2}{3}}+\sqrt{\frac{\left(y+2z\right)^2}{3}}+\sqrt{\frac{\left(z+2x\right)^2}{3}}\)
\(\Rightarrow P\ge\frac{1}{\sqrt{3}}\left(3x+3y+3z\right)\ge\frac{3}{\sqrt{3}}=\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)
a, b, c > 0. CM:
a)\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{\frac{a^2+b^2}{2}}+\sqrt{\frac{b^2+c^2}{2}}+\sqrt{\frac{c^2+a^2}{2}}\)
b)\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2+b^2-ab}+\sqrt{b^2+c^2-bc}+\sqrt{c^2+a^2-ac}\)
a/ \(\frac{b}{b}.\sqrt{\frac{a^2+b^2}{2}}+\frac{c}{c}.\sqrt{\frac{b^2+c^2}{2}}+\frac{a}{a}.\sqrt{\frac{c^2+a^2}{2}}\)
\(\le\frac{1}{b}.\left(\frac{3b^2+a^2}{4}\right)+\frac{1}{c}.\left(\frac{3c^2+b^2}{4}\right)+\frac{1}{a}.\left(\frac{3a^2+c^2}{4}\right)\)
\(=\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\)
Ta cần chứng minh
\(\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\left(a+b+c\right)\)
Mà: \(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Vậy có ĐPCM.
Câu b làm y chang.
Cho 3 số dương a;b;c thỏa mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=1\)
CMR: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2\sqrt{2}}\)
Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x,y,z>0\\x+y+z=1\end{cases}}\)
Và \(\hept{\begin{cases}a^2=\frac{x^2+z^2-y^2}{2}\\b^2=\frac{x^2+y^2-z^2}{2}\\c^2=\frac{y^2+z^2-x^2}{2}\end{cases}}\) và \(\hept{\begin{cases}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{cases}}\)
\(\Rightarrow VT\ge\frac{1}{2\sqrt{2}}\left(\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{2z}+\frac{y^2+z^2-x^2}{x}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{2\left(x+y+z\right)^2}{x+y+z}-\left(x+y+z\right)\right)\)
\(=\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2\sqrt{2}}\)
(4)Bài 1:Với \(\forall\) a>b>0. CMR: a+ \(\frac{1}{b\left(a-b\right)}\ge3\)
(7) Bài 2: Cho a,b,c \(\ne\) 0 .CMR: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
(8) Bài 3: Cho a,b,c>0 thõa mãn abc=1
CMR: \(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Bài 1: \(a+\frac{1}{b\left(a-b\right)}=\left(a-b\right)+b+\frac{1}{b\left(a-b\right)}\)
Áp dụng BĐT Cauchy cho 3 số dương ta thu được đpcm (mình làm ở đâu đó rồi mà:)
Dấu "=" xảy ra khi a =2; b =1 (tự giải ra)
Bài 2: Thêm đk a,b,c >0.
Theo BĐT Cauchy \(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{c^2}}=\frac{2a}{c}\). Tương tự với hai cặp còn lại và cộng theo vế ròi 6chia cho 2 hai có đpcm.
Bài 3: Nó sao sao ấy ta?