=(x²+2xy+y²)+(y²-4yz+4z²)+(y²-2y+1)+(z²-2z+1)-4x-2y-4z+5

=(x+y)²-4(x+y)+4 +(y-2z)²+2(y-2z)+1 +(y-1)²+(z-1)²

=(x+y-2)²+(y-2z+1)²+(y-1)²+(z-1)²\(\ge0\)\(\forall_{x,y,z}\)

Lai co (x+y-2)²+(y-2z+1)²+(y-1)²+(z-1)²\(\le\)0

=> (x+y-2)²+(y-2z+1)²+(y-1)²+(z-1)²=0

Dau = xay ra khi x=y=z=1

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{3}=\frac{y}{6}=\frac{z}{5}=\frac{2xy-3y^2-4yz}{2.3.6-3.6^2-4.6.5}=\frac{24}{-192}=\left(-\frac{1}{8}\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=\left(-\frac{1}{8}\right)\Rightarrow x=\left(-\frac{3}{8}\right)\\\frac{y}{6}=\left(-\frac{1}{8}\right)\Rightarrow y=\left(-\frac{3}{4}\right)\\\frac{z}{5}=\left(-\frac{1}{8}\right)\Rightarrow z=\left(-\frac{5}{8}\right)\end{cases}}\)

Vậy .... 

\(b,\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}=\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{2x+3y-5z}{2.2+3.3-5.5}=\frac{\left(-12\right)}{\left(-12\right)}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=1\Rightarrow x=2\\\frac{y}{3}=1\Rightarrow y=3\\\frac{z}{5}=1\Rightarrow z=5\end{cases}}\)

Vậy ... 

a/ a² + b² + c² \(\ge\)ab + bc + ca

<=> 2(a² + b² + c²⁾ \(\ge\)2(ab + bc + ca)

<=> (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a² \(\ge0\)

<=> (a - b)² + (b - c)² + (c - a)² \(\ge0\) (đúng)

=> ĐPCM

b/ a² + b² + c² \(\ge\) 2ab - 2ac + 2bc

<=> a² + b² + c² + 2( - ab + ac - bc)\(\ge\) 0

<=> (a - b + c)² \(\ge0\)(đúng)

=> ĐPCM

Câu hỏi của thanh ngọc - Toán lớp 9 | Học trực tuyến

1/

Xét hiệu $(x+1)^2-4x^2=(x+1)^2-(2x)^2=(x+1-2x)(x+1+2x)$

$=(1-x)(3x+1)$
Do $x\in (0;1)$ nên $1-x>0; 3x+1>0$

$\Rightarrow (x+1)^2-4x^2>0\Rightarrow (x+1)^2> 4x^2$

2/

Xét hiệu:

$(1+x+y)^2-4(x^2+y^2)=x^2+y^2+1+2x+2y+2xy-4x^2-4y^2$

$=1+2x+2y+2xy-3x^2-3y^2$

$=2x(1-x)+2y(1-y)+1+2xy-x^2-y^2$
Vì $x,y\in (0;1)$ nên: 

$2x(1-x)>0$

$2y(1-y)>0$

$(x-1)(y-1)>0\Rightarrow xy+1> x+y=x.1+y.1> x^2+y^2$

$\Rightarrow 1+xy-x^2-y^2>0$

$\Rightarrow 1+2xy-x^2-y^2>0$

Suy ra: $2x(1-x)+2y(1-y)+1+2xy-x^2-y^2>0$

$\Rightarrow (1+x+y)^2> 4(x^2+y^2)$

Lời giải:
\(\frac{4x-5y}{3}=\frac{5z-3x}{4}=\frac{3y-4z}{5}\)

\(=\frac{3(4x-5y)}{9}=\frac{4(5z-3x)}{16}=\frac{5(3y-4z)}{25}\)

\(=\frac{12x-15y}{9}=\frac{20z-12x}{16}=\frac{15y-20z}{25}=\frac{12x-15y+20z-12x+15y-20z}{9+16+25}=0\)

\(\Rightarrow 4x-5y=5z-3x=3y-4z=0\)

\(\Rightarrow 4x=5y; 3y=4z\Rightarrow \frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)