dùng hệ số bất định ấy ,lười lắm

p. tích thành tổng 2 bình phương rồi mincopxki

Dễ chứng minh được \(2x^2+3xy+2y^2\ge\frac{7}{4}\left(x+y\right)^2\)

                       \(\Leftrightarrow\left(\frac{1}{2}x-\frac{1}{2}y\right)^2\ge0\left(true\right)\)

Một cách tương tự :

\(2y^2+3yz+2z^2\ge\frac{7}{4}\left(y+z\right)^2\)

\(2z^2+3xz+2x^2\ge\frac{7}{4}\left(z+x\right)^2\)

\(\Rightarrow A=\sqrt{2x^2+3xy+2y^2}+\sqrt{2y^2+3yz+2z^2}+\sqrt{2z^2+3xz+2x^2}\)

\(\ge\sqrt{\frac{7}{4}\left(x+y\right)^2}+\sqrt{\frac{7}{4}\left(y+z\right)^2}+\sqrt{\frac{7}{4}\left(z+x\right)^2}\)

\(=\frac{\sqrt{7}}{2}\left(x+y+y+z+z+x\right)=\frac{\sqrt{7}}{2}.6=3\sqrt{7}\)

Lời giải:

Vì $x+y+z=0$ nên $x+y=-z$. Do đó:

$M=2(x^3+y^3)+2z(z^2-3xy)$

$=2[(x+y)^3-3xy(x+y)]+2z^3-6xyz$

$=2[(-z)^3+3xyz]+2z^3-6xyz=-2z^3+6xyz+2z^3-6xyz=0$

Lời giải:

\(\frac{1}{x^2}=1-\frac{1}{y^2}-\frac{1}{z^2}<1\Rightarrow x^2-1>0\)

\(P=\frac{y^2z^2}{x(y^2+z^2)}+\frac{x^2z^2}{y(x^2+z^2)}+\frac{x^2y^2}{z(x^2+y^2)}\)

\(=\frac{1}{x(\frac{1}{y^2}+\frac{1}{z^2})}+\frac{1}{y(\frac{1}{x^2}+\frac{1}{z^2})}+\frac{1}{z(\frac{1}{x^2}+\frac{1}{y^2})}\)

\(=\frac{1}{x(1-\frac{1}{x^2})}+\frac{1}{y(1-\frac{1}{y^2})}+\frac{1}{z(1-\frac{1}{z^2})}\)

\(=\frac{x}{x^2-1}+\frac{y}{y^2-1}+\frac{z}{z^2-1}\)

Xét đánh giá sau:

\(\frac{x}{x^2-1}-\frac{3\sqrt{3}}{2x^2}=\frac{(x-\sqrt{3})^2(2x+\sqrt{3})}{2x^2(x^2-1)}\geq 0, \forall x^2>1\)

\(\Rightarrow \frac{x}{x^2-1}\geq \frac{3\sqrt{3}}{2x^2}\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:

\(\Rightarrow P=\frac{x}{x^2-1}+\frac{y}{y^2-1}+\frac{z}{z^2-1}\geq \frac{3\sqrt{3}}{2}(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\frac{3\sqrt{3}}{2}\)

Vậy \(P_{\min}=\frac{3\sqrt{3}}{2}\Leftrightarrow x=y=z=\sqrt{3}\)

SOS get it <(")

\(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)->\left(a;;bc\right)\text{for}\left(a;b;c>0\text{and}a^2+b^2+c^2=1\right)\)

\(\text{Khido}P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)

\(\text{Ta se cm}\sum_{cyc}\frac{a}{b^2+c^2}\ge\frac{3\sqrt{3}}{2}\)\(\text{Viet lai BDT can chung minh}\)

\(\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}}{2\sqrt{x^2+y^2+z^2}}\)

\(\text{Chuan hoa}a^2+b^2+c^2=3\text{ta can cm:}\)

\(\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{3-a^2}+\frac{b}{3-b^2}+\frac{c}{3-c^2}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{3-a^2}-\frac{1}{2}+\frac{b}{3-b^2}-\frac{1}{2}+\frac{c}{3-c^2}-\frac{1}{2}\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\frac{a}{3-a^2}-\frac{1}{2}-\frac{1}{2}\left(x^2-1\right)\right)\ge0\)

\(\Leftrightarrow\frac{a\left(a+2\right)\left(a-1\right)^2}{3-a^2}+\frac{b\left(b+2\right)\left(b-1\right)^2}{3-b^2}+\frac{c\left(c+2\right)\left(c-1\right)^2}{3-c^2}\ge0\)

C/m: \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

\(\Rightarrow2x^2+xy+2y^2\ge\dfrac{5}{4}\left(x^2+2xy+y^2\right)\)

\(\Leftrightarrow8x^2+4xy+8y^2\ge5x^2+10xy+5y^2\)

\(\Leftrightarrow3\left(x-y\right)^2\ge0\left(LĐ\right)\)

Vậy \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

CMTT: \(\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)\);

\(\sqrt{2z^2+zx+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)

Vậy H=\(\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+xz+2z^2}\ge\sqrt{5}\left(x+y+z\right)=2019\)H_min=2019\(\Leftrightarrow x=y=z=\dfrac{\dfrac{2019}{\sqrt{5}}}{3}\)

Khos quas

Ta có: \(x^2-2y=-1\) \(\Leftrightarrow\) \(x^2-2y+1=0\) (1)

\(y^2+1=2z\) \(\Leftrightarrow y^2-2z+1=0\) (2)

\(2z^2=4x-2\) \(\Leftrightarrow2z^2-4x+2=0\)(3)

Cộng (1)(2)(3) theo vế:

\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)

=> x-1=0; y-1=0; z-1=0

=>x=y=z

=>\(x^{2015}+y^{2015}+z^{2015}=1+1+1=3\)(đpcm)

x^2-2y=-1=>x^2-2y+1=0

y^2+1=2z=>y^2-2z+1=0

2z^2=4x-2=>z^2-2x+1=0

cộng vế với vế của 3 pt

ta có x^2-2y+1+y^2-2z+1+z^2-2x+1=0

=>(x-1)^2+(y-1)^2+(z-1)^2=0

=>x-1=0; y-1=0; z-1=0;

=>x=y=z=1

=>x^2015+y^2015+z^2015=3