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![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mk muốn làm giúp bạn lắm chứ nhưng mà khổ lỗi mk mới học lớp 6 . Xin lỗi bn
bài 2 gợi ý từ hdt (x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)
VT (ở đề bài) = a+b+c
<=>....<=>3[căn bậc 3(a)+căn bậc 3(b)].[căn bậc 3(b)+căn bậc 3(c)].[căn bậc 3(c)+căn bậc 3 (a)]=0
từ đây rút a=-b,b=-c,c=-a đến đây tự giải quyết đc r
![](https://rs.olm.vn/images/avt/0.png?1311)
bài n t vừa làm mà, vào link này nhé
https://olm.vn/hoi-dap/question/1129328.html
![](https://rs.olm.vn/images/avt/0.png?1311)
vì 0<a<1 ;0<b<2 ;0<c<3
=> 1-a > 0 <=> 0<\(\sqrt{1-a}\) < 1
=> 0 <\(\dfrac{\sqrt{1-a}}{a}\) ≤ 1 (1)
c/m tương tự với b,c
=> 0 < \(\dfrac{\sqrt{2-b}}{b}\) ≤ 2 (2)
và 0 < \(\dfrac{\sqrt{3-c}}{c}\) ≤ 3 (3)
Cộng các vế của bđt với nhau
=> 0 < \(\dfrac{\sqrt{1-a}}{a}+\dfrac{\sqrt{2-b}}{b}+\dfrac{\sqrt{3-c}}{c}\) ≤ 6
Vậy GTLN của A là 6
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
Với mọi số thực dương x;y;z ta có:
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
\(\Leftrightarrow3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2xy+2yz+2zx\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
Áp dụng:
a.
\(\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\le\sqrt{3\left(a+2+b+2+c+2\right)}=\sqrt{3\left(21+6\right)}=9\)
b.
\(\sqrt{a+b+2}+\sqrt{b+c+2}+\sqrt{c+a+2}\le\sqrt{3\left(a+b+2+b+c+2+c+a+2\right)}\)
\(\Rightarrow\sqrt{a+b+2}+\sqrt{b+c+2}+\sqrt{c+a+2}\le\sqrt{6\left(a+b+c\right)+18}=\sqrt{6.21+18}=12\)
Dấu "=" xảy ra khi \(a=b=c=7\)