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: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 3 2022
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 \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 10 2021
\(P\sqrt{2}\ge\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{c^2+a^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\)
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
28 tháng 1 2018
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 \(\left\{{}\begin{matrix}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{matrix}\right.\)
\(VT=\dfrac{1}{2\sqrt{2}}\left(\dfrac{x^2+z^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}+\dfrac{y^2+z^2-x^2}{x}\right)\)
\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)}-\left(x+y+z\right)\right)\)
\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{\sqrt{2011}}{2\sqrt{2}}=VP\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2022
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
2 tháng 1 2022
Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)
Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)
\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)
\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
\(b+c\le\sqrt{2\left(b^2+c^2\right)}\Rightarrow\dfrac{a^2}{b+c}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{1}{\sqrt{2}}.\dfrac{a^2}{\sqrt{b^2+c^2}}\)
Sau đó làm tiếp như bài đó là được
Nguyễn Việt Lâm Giáo viên, áp dụng BĐT gì vậy bn??