`x+y+z>=0` là chưa đủ phải là `x,y,z>=0` mới đúng.

`x+y+z>=sqrt{xy}+sqrt{yz}+sqrt{zx}`

`<=>2x+2y+2z>=2sqrt{xy}+2sqrt{yz}+2sqrt{zx}`

`<=>x-2sqrt{xy}+y+y-2sqrt{yz}+z+z-2sqrt{zx}+x>=0`

`<=>(sqrtx-sqrty)^2+(sqrty-sqrtz)^2+(sqrtz-sqrtx)^2>=0` luôn đúng

Dấu `"="<=>x=y=z`

Áp dụng bdt Co-si, ta có:

\(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(z+x\ge2\sqrt{xz}\)

=> 2(x+y+z) \(\ge2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

=> đpcm

Bạn tham khảo lời giải tại đây:

cho các số thực dưong x,y,z thỏa mãn : x2 y2 z2=3chứng minh rằng : \(\dfrac{x}{\sqrt[3]{yz}} \dfrac{y}{\sqrt[3]{zx}} \df... - Hoc24

Cách khác:

Áp dụng BĐT AM-GM và BĐT Cauchy-Schwarz:

\(\sum \frac{x}{\sqrt[3]{yz}}\geq \sum \frac{x}{\frac{y+z+1}{3}}=3\sum \frac{x}{y+z+1}=3\sum \frac{x^2}{xy+xz+x}\)

\(\geq 3. \frac{(x+y+z)^2}{2(xy+yz+xz)+(x+y+z)}\)

Ta sẽ chứng minh: \(\frac{3(x+y+z)^2}{2(xy+yz+xz)+(x+y+z)}\geq xy+yz+xz(*)\)

Đặt $x+y+z=a$ thì $xy+yz+xz=\frac{a^2-3}{2}$

Bằng BĐT AM-GM dễ thấy $\sqrt{3}< a\leq 3$

BĐT $(*)$ trở thành:

$\frac{3a^2}{a^2+a-3}\geq \frac{a^2-3}{2}$

$\Leftrightarrow a^4+a^3-12a^2-3a+9\leq 0$

$\Leftrightarrow (a-3)(a+1)(a^2+3a-3)\leq 0$

Điều này đúng với mọi $\sqrt{3}< a\leq 3$

Do đó BĐT $(*)$ đúng nên ta có đpcm.

Dấu "=" xảy ra khi $x=y=z=1$

\(x^2-xy+y^2=\dfrac{1}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)

\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\sqrt{\dfrac{1}{4}\left(x+y\right)^2}=\dfrac{1}{2}\left(x+y\right)\)

Tương tự: \(\sqrt{y^2-yz+z^2}\ge\dfrac{1}{2}\left(y+z\right)\); \(\sqrt{z^2-zx+x^2}\ge\dfrac{1}{2}\left(z+x\right)\)

Cộng vế:

\(Q\ge\dfrac{1}{2}\left(x+y\right)+\dfrac{1}{2}\left(y+z\right)+\dfrac{1}{2}\left(z+x\right)=x+y+z=3\) (đpcm)

Ta có : Áp dụng BĐT Cauchy ba số ở mẫu ta được

\(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}=\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\)Thấy: \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}\left(?!\right)\)

Ta phải chứng minh:

\(\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{3}\)

\(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{9}\)

Mà \(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}=\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\)

Theo C.B.S

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

Phải chứng minh

\(\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{9}\)

\(\Leftrightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)

Ta có : \(xy+yz+xz\le x^2+y^2+z^2=3\)

Theo C.B.S : \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)

\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le9\)

\(\Rightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)

=> ĐPCM

ta sử dụng bđt :\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)(dk mọi abcd)
cái này cm dễ thôi. bunhia nha
ĐĂT :\(A=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)
\(\Rightarrow A=\sqrt{\left(x+\frac{y}{2}\right)^2+\left(\frac{y\sqrt{3}}{2}\right)^2}+\sqrt{\left(y+\frac{z}{2}\right)^2+\left(\frac{z\sqrt{3}}{2}\right)^2}+\sqrt{\left(z+\frac{x}{2}\right)^2+\left(\frac{x\sqrt{3}}{2}\right)^2}\)
Áp dingj bđt trên ta được \(A\ge\sqrt{\left(x+\frac{y}{2}+y+\frac{z}{2}+z+\frac{x}{2}\right)^2+\left(\frac{x\sqrt{3}}{2}+\frac{y\sqrt{3}}{2}+\frac{z\sqrt{3}}{2}\right)^2}\)
\(\Rightarrow A\ge\sqrt{\frac{9}{4}\left(x+y+z\right)^2+\frac{3}{4}\left(x+y+z\right)^2}=\sqrt{3}\left(x+y+z\right)\)(dpcm)
Dấu = xảy ra khi và chỉ khi x=y=z

\(\sqrt{x^2+xy+y^2}=\sqrt{\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2}\ge\sqrt{\frac{3}{4}\left(x+y\right)^2}=\frac{\sqrt{3}}{2}\left(x+y\right)\)

cách khác

ÁP DỤNG BĐT Mincopxki

\(VT=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\)

\(=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}}+\sqrt{\left(y+\frac{z}{2}\right)^2+\frac{3z^2}{4}}+\sqrt{\left(x+\frac{z}{2}\right)^2+\frac{3z^2}{4}}\)

\(\ge\sqrt{\left(x+y+z+\frac{x+y+z}{2}\right)^2+\left(\frac{\sqrt{3}\left(x+y+z\right)}{2}\right)^2}\)

\(=\sqrt{\frac{9\left(x+y+z\right)^2}{4}+\frac{3\left(x+y+z\right)^2}{4}}\)

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

Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)

Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)

\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)

Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)

\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)

\(\Rightarrow A\ge\dfrac{1}{2}\)

Biến đổi tương đương là ok mà

Ta có; \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)

<=> \(2x+2y+2z-2\sqrt{xy}-2\sqrt{yz}-2\sqrt{xz}\ge0\)

<=> \(\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{xz}+x\right)\ge0\)

<=> \(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{x}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\)

( Luôn đúng)

=> đpcm

Dấu = xảy ra <=> \(x=y=z\)