Bài này hôm trước hình như bạn mới hỏi xong, vậy làm chi tiết cho đỡ băn khoăn:

Với các số dương a;b;c;x;y;z bất kì, ta chứng minh BĐT sau:

\(\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge\sqrt{\left(a+b\right)^2+\left(x+y\right)^2}\)

Thật vậy, BĐT tương đương:

\(a^2+b^2+x^2+y^2+2\sqrt{a^2b^2+x^2y^2+x^2b^2+a^2y^2}\ge a^2+b^2+x^2+y^2+2ab+2xy\)

\(\Leftrightarrow\sqrt{a^2b^2+x^2y^2+a^2y^2+b^2x^2}\ge ab+xy\)

\(\Leftrightarrow a^2b^2+x^2y^2+a^2y^2+b^2x^2\ge a^2b^2+x^2y^2+2abxy\)

\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) (luôn đúng)

Từ đó suy ra:

\(\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\ge\sqrt{\left(a+b\right)^2+\left(x+y\right)^2}+\sqrt{c^2+z^2}\ge\sqrt{\left(a+b+c\right)^2+\left(x+y+z\right)^2}\)

Áp dụng cho bài toán:

\(VT=\sqrt{\left(x+\dfrac{y}{2}\right)^2+\left(\dfrac{\sqrt{3}y}{2}\right)^2}+\sqrt{\left(y+\dfrac{z}{2}\right)^2+\left(\dfrac{\sqrt{3}z}{2}\right)^2}+\sqrt{\left(z+\dfrac{x}{2}\right)^2+\left(\dfrac{\sqrt{3}x}{2}\right)^2}\)

\(VT\ge\sqrt{\left(x+\dfrac{y}{2}+y+\dfrac{z}{2}+z+\dfrac{x}{2}\right)^2+\left(\dfrac{\sqrt{3}y}{2}+\dfrac{\sqrt{3}z}{2}+\dfrac{\sqrt{3}x}{2}\right)^2}=2\left(x+y+z\right)\) (đpcm)

\(\Leftrightarrow\sqrt{4x^2+4xy+8y^2}+\sqrt{4y^2+4yz+8z^2}+\sqrt{4z^2+4zx+8x^2}\ge4\left(x+y+z\right)\)

Ta có:

\(VT=\sqrt{\left(2x+y\right)^2+\left(\sqrt{7}y\right)^2}+\sqrt{\left(2y+z\right)^2+\left(\sqrt{7}z\right)^2}+\sqrt{\left(2z+x\right)^2+\left(\sqrt{7}x\right)^2}\)

\(VT\ge\sqrt{\left(2x+y+2y+z+2z+x\right)^2+\left(\sqrt{7}x+\sqrt{7}y+\sqrt{7}z\right)^2}\)

\(VT\ge\sqrt{16\left(x+y+z\right)^2}=4\left(x+y+z\right)\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\)

ai giúp mình với

...

Ta có:
         \(\dfrac{x-y}{1+xy}\)+\(\dfrac{y-z}{1+yz}\)+\(\dfrac{z-x}{1+xz}\) = \(\dfrac{x-y}{1+xy}\)+\(\dfrac{-\left(x-y\right)-\left(z-x\right)}{1+yz}\)+\(\dfrac{z-x}{1+xz}\)

         =\(\dfrac{x-y}{1+xy}\)\(-\dfrac{x-y}{1+yz}\) \(-\dfrac{z-x}{1+yz}\)+\(\dfrac{z-x}{1+xz}\) 

         = \(\left(x-y\right)\)\(\left(\dfrac{\left(1+yz\right)-\left(1+xy\right)}{\left(1+yz\right)\left(1+xy\right)}\right)\)+(\(z-x\))\(\left(\dfrac{\left(1+yz\right)-\left(1+zx\right)}{\left(1+yz\right)\left(1+zx\right)}\right)\)

         =\(\left(x-y\right)\)\(\dfrac{y\left(z-x\right)}{\left(1+yz\right)\left(1+xy\right)}\)+(\(z-x\))\(\dfrac{-z\left(x-y\right)}{\left(1+yz\right)\left(1+zx\right)}\)

         =\(\left(\dfrac{\left(x-y\right)\left(z-x\right)}{1+yz}\right)\)\(\left(\dfrac{y\left(1+xz\right)-z\left(1+xy\right)}{\left(1+xz\right)\left(1+xy\right)}\right)\)

       =đpcm

Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(y+z\right)\sqrt{yz}}{x}\ge\frac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\frac{2\sqrt{\left(yz\right)^2}}{x}=\frac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại ta cũng có

\(\frac{\left(x+y\right)\sqrt{xy}}{z}\ge\frac{2xy}{z};\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xz}{y}\)

\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{yz}}{x}+\frac{\left(x+y\right)\sqrt{xy}}{z}+\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\)

Cần chứng minh \(\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge x+y+z\)

Áp dụng BĐT AM-GM:

\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế ta có ĐPCM

Khi \(x=y=z\)

\(x^2+y^2+z^2=xy+yz+zx\)

=> \(2x^2+2y^2+2x^2=2xy+2yz+2zx\) 

=> \(2x^2+2y^2+2x^2-2xy-2yz-2zx=0\) 

=> \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\) 

=> x -y =0 ; y - z=0 ; z - x=0

=> x =y; y =z; z=x

=> x=y=z

Lời giải:

Đặt \((x,y,z)=(a^2,b^2,c^2)\). Bài toán tương đương với:

\(\frac{bc(b+c)}{a}+\frac{ac(a+c)}{b}+\frac{ab(a+b)}{c}\geq 2(a^2+b^2+c^2)\)

Biến đổi ta thấy:

\(\text{VT}=a^2\left ( \frac{b}{c}+\frac{c}{b} \right )+b^2\left ( \frac{a}{c}+\frac{c}{a} \right )+c^2\left ( \frac{a}{b}+\frac{b}{a} \right )\)

Áp dụng BĐT AM-GM:

\(\left\{\begin{matrix} \frac{a}{b}+\frac{b}{a}\geq 2\\ \frac{a}{c}+\frac{c}{a}\geq 2\\ \frac{b}{c}+\frac{c}{b}\geq 2\end{matrix}\right.\Rightarrow \text{VT}\geq 2(a^2+b^2+c^2)=\text{VP}\)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\Leftrightarrow x=y=z>0\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{\left(y+z\right)\sqrt{yz}}{x}\ge\dfrac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\dfrac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại thì được:

\(\dfrac{2xy}{z}+\dfrac{2yz}{x}+\dfrac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge x+y+z\)

Tiếp tục dùng AM-GM:

\(\dfrac{xy}{z}+\dfrac{yz}{x}\ge2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế có:

\(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge x+y+z\) (đúng)

Hay ta có ĐPCM. Khi \(x=y=z\)

Đề này à: \(\dfrac{\left(y+z\right)\sqrt{yz}}{x}+\dfrac{\left(z+x\right)\sqrt{zx}}{y}+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge2\left(x+y+z\right)\)

Dùng máy tính kiểm tra. (đề sai không?)

Thế x=1, y=2, z=3

VT = 17,12576389

VP = 12

Lời giải:

$2\text{VT}=2(x+y+z)-4(xy+yz+xz)+8xyz$

$=(2x-1)(2y-1)(2z-1)+1$

Do $x,y,z\in [0;1]$ nên $-1\leq 2x-1, 2y-1, 2z-1\leq 1$

$\Rightarrow (2x-1)(2y-1)(2z-1)\leq 1$

$\Rightarrow 2\text{VT}\leq 2$

$\Rightarrow \text{VT}\leq 1$
Ta có đpcm.

Dấu "=" xảy ra khi $(x,y,z)=(1,1,1), (0,0,1)$ và hoán vị.