Cho a,b,c >0 t/m \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\).
TÌm min \(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
Cho a, b, c > 0 thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}=\sqrt{2011}\). C\m :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}.\sqrt{\dfrac{2011}{2}}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{x^2+z^2-y^2}{2}\\b^2=\dfrac{x^2+y^2-z^2}{2}\\c^2=\dfrac{y^2+z^2-x^2}{2}\\x+y+z=\sqrt{2011}\end{matrix}\right.\)
Và \(\left\{{}\begin{matrix}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{matrix}\right.\)
\(VT=\dfrac{1}{2\sqrt{2}}\left(\dfrac{x^2+z^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}+\dfrac{y^2+z^2-x^2}{x}\right)\)
\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)}-\left(x+y+z\right)\right)\)
\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{\sqrt{2011}}{2\sqrt{2}}=VP\)
cho a,b,c>0 thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=3\sqrt{2}\)
tìm min của \(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{a^2+b^2}=z\\\sqrt{b^2+c^2}=x\\\sqrt{c^2+a^2}=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=z^2\\b^2+c^2=x^2\\c^2+a^2=y^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
Điều kiện đề bài thành: \(x+y+z=3\sqrt{2}\)
Ta có:
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
\(=\dfrac{y^2+z^2-x^2}{2\sqrt{2}x}+\dfrac{z^2+x^2-y^2}{2\sqrt{2}y}+\dfrac{x^2+y^2-z^2}{2\sqrt{2}z}\)
\(=\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-x-y-z\right)\)
\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}-x-y-z\right)\)
\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2\sqrt{2}}.3\sqrt{2}=\dfrac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{2}\) hay \(a=b=c=1\)
cho a,b,c dương thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\).
CMR: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\sqrt{2011}}{2}\)
Đặt \(\left(x,y,z\right)=\left(\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+a^2}\right)\).
Ta có \(x+y+z=\sqrt{2011}\).
BĐT cần cm trở thành:
\(\dfrac{y^2+z^2-x^2}{2\sqrt{2}x}+\dfrac{z^2+x^2-y^2}{2\sqrt{2}y}+\dfrac{x^2+y^2-z^2}{2\sqrt{2}z}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
\(\Leftrightarrow\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\ge x+y+z\)
\(\Leftrightarrow\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{y}\right)+\left(\dfrac{y^2}{x}+\dfrac{z^2}{y}+\dfrac{x^2}{z}\right)\ge2\left(x+y+z\right)\).
Theo bđt AM - GM:
\(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{y}=\left(\dfrac{x^2}{y}+y\right)+\left(\dfrac{y^2}{z}+z\right)+\left(\dfrac{z^2}{x}+x\right)-x-y-z\ge2x+2y+2z-x-y-z=x+y+z\).
Tương tự, \(\dfrac{y^2}{x}+\dfrac{z^2}{y}+\dfrac{x^2}{z}\ge x+y+z\).
Dễ có điều phải chứng minh.
\(P\sqrt{2}\ge\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{c^2+a^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2011}\\a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow P2\sqrt{2}\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)
\(P4\sqrt{2}\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)
\(P2\sqrt{2}\ge\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}-\left(x+y+z\right)=x+y+z=\sqrt{2011}\)
\(\Rightarrow P\ge\dfrac{\sqrt{2011}}{2\sqrt{2}}\)
Đề sai
Cho a;b;c;d>0 thỏa mãn: a+b+c+d=4. Tìm min của:
\(\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{d^2}}+\sqrt{d^2+\dfrac{1}{a^2}}\)
Áp dụng bđt Mincopxki và Cauchy-Schwarz:
\(VT=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{d^2}}+\sqrt{d^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c+d\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2}\)
\(\ge\sqrt{\left(a+b+c+d\right)^2+\left(\dfrac{16}{a+b+c+d}\right)^2}\)
\(=\sqrt{3^2+\dfrac{16^2}{3^2}}=\sqrt{\dfrac{337}{9}}\)
\("="\Leftrightarrow a=b=c=d=\dfrac{3}{4}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(a^2+\frac{1}{b^2}\right)(1+1)\geq (a+\frac{1}{b})^2\)
\(\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{a+\frac{1}{b}}{\sqrt{2}}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{d^2}}+\sqrt{d^2+\frac{1}{a^2}}\geq \frac{1}{\sqrt{2}}(a+b+c+d+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\)
Mặt khác theo BĐT Cauchy:
\(a+\frac{1}{a}\geq 2; b+\frac{1}{b}\geq 2; c+\frac{1}{c}\geq 2; d+\frac{1}{d}\geq 2\)
\(\Rightarrow \text{VT}\geq \frac{1}{\sqrt{2}}.8=4\sqrt{2}\)
Vậy giá trị nhỏ nhất của biểu thức là $4\sqrt{2}$. Dấu bằng xảy ra khi $a=b=c=d=1$
Cho 3 số dương a,b,c thỏa mãn
\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\)
CMR:\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Đặt vế trái BĐT cần chứng minh là P
Ta có:
\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2011}\)
Đồng thời: \(\left\{{}\begin{matrix}y^2+z^2-x^2=2a^2\\z^2+x^2-y^2=2b^2\\x^2+y^2-z^2=2c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z+z+x+x+y\right)^2}{2x+2y+2z}-\left(x+y+z\right)\right)=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)
Cho các số dương a, b, c thỏa mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\). CMR: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}++\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Cho các số dương a, b, c thỏa mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\). CMR: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
\(b+c\le\sqrt{2\left(b^2+c^2\right)}\Rightarrow\dfrac{a^2}{b+c}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{1}{\sqrt{2}}.\dfrac{a^2}{\sqrt{b^2+c^2}}\)
Sau đó làm tiếp như bài đó là được
Cho 3 số dương a, b, c thoả mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\). Chứng minh rằng: \(\dfrac{a^2}{b+c}+\dfrac{a^2}{b+c}+\dfrac{a^2}{b+c}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Cho a,b,c >0 thỏa a+b+c \(\ge9\)
Tìm Min:
\(P=2\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}+\sqrt{\dfrac{1}{a}+\dfrac{9}{b}+\dfrac{25}{c}}\)
cái kia là \(3\sqrt{\dfrac{1}{a}+\dfrac{9}{b}+\dfrac{25}{c}}\)
\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)