![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng BĐT AM-GM:
$1=xy+yz+xz+2xyz\leq \frac{(x+y+z)^2}{3}+2.\frac{(x+y+z)^3}{27}$
$\Leftrightarrow 1\leq \frac{t^2}{3}+\frac{2t^3}{27}$ (đặt $x+y+z=t$)
$\Leftrightarrow 2t^3+9t^2-27\geq 0$
$\Leftrightarrow (t+3)^2(2t-3)\geq 0$
$\Leftrightarrow 2t-3\geq 0$
$\Leftrightarrow t\geq \frac{3}{2}$ hay $x+y+z\geq \frac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $x=y=z=\frac{1}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{x+y}{xy}.\frac{1}{z}\ge\frac{2\sqrt{xy}}{xy}.\frac{1}{z}=\frac{2}{z\sqrt{xy}}\ge\frac{2}{z\left(\frac{x+y}{2}\right)}=\frac{4}{z\left(x+y\right)}\)
\(=\frac{4}{z\left(1-z\right)}=\frac{4}{\frac{1}{4}-\left(z-\frac{1}{2}\right)^2}\ge16\)
Min M= 16 khi z=1/2 và x=y =1/4.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Côsi cho 2 số dương x và \(\sqrt{1-y^2}\) có:
x\(\sqrt{1-y^2}\) ≤ \(\dfrac{x^2+1-y^2}{2}\)
Tương tự: \(y\sqrt{1-z^2}\le\dfrac{y^2+1-z^2}{2}\); \(z\sqrt{1-x^2}\le\dfrac{z^2+1-x^2}{2}\)
=> \(x\sqrt{1-y^2}+y\sqrt{1-z^2}+z\sqrt{1-x^2}\le\dfrac{x^2+1-y^2+y^2+1-z^2+z^2+1-x^2}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra ⇔ x = y = z = \(\dfrac{\sqrt{2}}{2}\) => x2 = y2 = z2 = \(\dfrac{1}{2}\)
=> x2 + y2 + z2 = 3x2 = 3.\(\dfrac{1}{2}\) = \(\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)
\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)
Tương tự cũng có:
\(\sqrt{\dfrac{yz}{y+z+2x}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)\)
\(\sqrt{\dfrac{zx}{z+x+2y}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
Cộng vế với vế ta được:
\(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)