Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b+c\le\sqrt{2\left(b^2+c^2\right)}\Rightarrow\dfrac{a^2}{b+c}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{1}{\sqrt{2}}.\dfrac{a^2}{\sqrt{b^2+c^2}}\)
Sau đó làm tiếp như bài đó là được
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\sqrt{2}\ge\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{c^2+a^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Tuogw tựCâu hỏi của Nue nguyen - Toán lớp 10 | Học trực tuyến
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng bđt bunhia copxki ta có:
\(\sqrt{\left(1+1\right)\left(a^2+b^2\right)}=\sqrt{2\left(a^2+b^2\right)}>=\sqrt{\left(a+b\right)^2}=a+b\Rightarrow\sqrt{a^2+b^2}>=\frac{a+b}{\sqrt{2}}\)
\(\sqrt{\left(1+1\right)\left(b^2+c^2\right)}=\sqrt{2\left(b^2+c^2\right)}>=\sqrt{\left(b+c\right)^2}=b+c\Rightarrow\sqrt{b^2+c^2}>=\frac{b+c}{\sqrt{2}}\)
\(\sqrt{\left(1+1\right)\left(a^2+c^2\right)}=\sqrt{2\left(a^2+c^2\right)}>=\sqrt{\left(a+c\right)^2}=a+c\Rightarrow\sqrt{a^2+c^2}>=\frac{a+c}{\sqrt{2}}\)
\(\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}>=\frac{a+b+b+c+a+c}{\sqrt{2}}\)
\(=\frac{2\left(a+b+c\right)}{\sqrt{2}}=\frac{\left(\sqrt{2}\right)^2\left(a+b+c\right)}{\sqrt{2}}=\sqrt{2}\left(a+b+c\right)\)(đpcm)
dấu = xảy ra khi a=b=c
![](https://rs.olm.vn/images/avt/0.png?1311)
Do cả 2 vế cùng dương,ta bình phương 2 vế:
\(bđt\Leftrightarrow a^2+1+b^2+1+c^2+1+2\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+2\sqrt{\left(1+b^2\right)\left(1+c^2\right)}+2\sqrt{\left(1+c^2\right)\left(1+a^2\right)}\ge2\left(a+b+c\right)+2\sqrt{\left(a+b\right)\left(b+c\right)}+2\sqrt{\left(b+c\right)\left(c+a\right)}+2\sqrt{\left(c+a\right)\left(a+b\right)}\)Ta chứng minh từng bđt:
\(a^2+1+b^2+1+c^2+1\ge2\left(a+b+c\right)\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)(1)
Cần cm: \(\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+\sqrt{\left(1+b^2\right)\left(1+c^2\right)}+\sqrt{\left(1+c^2\right)\left(1+a^2\right)}\ge\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(c+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\)
Thật vậy theo Bunyakovsky: \(\sqrt{\left(1+a^2\right)\left(b^2+1\right)}+\sqrt{\left(1+b^2\right)\left(c^2+1\right)}+\sqrt{\left(1+c^2\right)\left(a^2+1\right)}\ge a+b+b+c+c+a=2\left(a+b+c\right)\)
Theo AM-GM: \(\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(a+c\right)}+\sqrt{\left(a+c\right)\left(a+b\right)}\le\frac{a+b+b+c+b+c+a+c+a+c+a+b}{2}=2\left(a+b+c\right)\) (2)
TTừ 1;2 ta có bđt được cm
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\hept{\begin{cases}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{cases}}\)
\(\Rightarrow VT\ge\frac{1}{2\sqrt{2}}\left(\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{2z}+\frac{y^2+z^2-x^2}{x}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{2\left(x+y+z\right)^2}{x+y+z}-\left(x+y+z\right)\right)\)
\(=\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(a=b=c=\frac{1}{3}\sqrt{\frac{2011}{2}}\)
Ta có : \(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\Rightarrow\sqrt{a^2+b^2}\ge\frac{\sqrt{2}}{2}\left(a+b\right)\)
\(\sqrt{b^2+c^2}\ge\frac{\sqrt{2}}{2}\left(b+c\right)\)
\(\sqrt{c^2+a^2}\ge\frac{\sqrt{2}}{2}\left(c+a\right)\)
\(\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\frac{\sqrt{2}}{2}.2.\left(a+b+c\right)=\sqrt{2}\)
@@ minh cung moi tim ra huong giai nhung chua hieu cach giai cua ban