Question

1. Chứng minh rằng:

\(x^2+y^2+z^2+3\ge2\cdot\left(x+y+z\right)\)

2. Cho a,b,c,d,e là các số thực, chứng minh rằng:

a) \(a^2+b^2+1\ge a\cdot b+a+b\)

b) \(a^2+b^2+c^2+d^2+e^2\ge a\cdot\left(b+c+d+e\right)\)

3. Cho a,b,c thỏa mãn:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Tính giá trị biểu thức: \(A=\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)\)

4. Tìm giá trị nhỏ nhất của biểu thức sau:

a) \(A=x\left(x-3\right)\left(x-4\right)\left(x-7\right)\)

b) \(A=\dfrac{3x^2-8x+6}{x^2-2x+1}\)

5. Cho \(x+y+z=3\)

a) Tìm GTNN của \(A=x^2+y^2+z^2\)

b) Tìm GTLN của \(B=xy+yz+xz\)

Answer 1

1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)

Từ (1), (2) và (3) suy ra:

\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)

<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm

Answer 2

5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) ⁽¹⁾

\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) ⁽²⁾

\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) ⁽³⁾

Từ (1),(2) và (3) suy ra:

\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)

<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)

mà x+y+z=3

=>\(x^2+y^2+z^2+3\ge2.3=6\)

<=> \(x^2+y^2+z^2\ge6-3=3\)

<=> \(A\ge3\)

Dấu "=" xảy ra khi x=y=z=1

Vậy GTNN của A=x²+y²+z² là 3 khi x=y=z=1

b, Ta có: x+y+z=3

=> \(\left(x+y+z\right)^2=9\)

<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)

<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)

mà \(x^2+y^2+z^2\ge3\) (theo a)

=> \(9-2xy-2yz-2xz\ge3\)

<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)

<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)

<=> \(B\le3\)

Dấu "=" xảy ra khi x=y=z=1

Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1

Answer 3

2, a, Ta có:

\(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\) (1)

\(\left(a-1\right)^2\ge0\Leftrightarrow a^2-2a+1\ge0\Leftrightarrow a^2+1\ge2a\) (2)

\(\left(b-1\right)^2\ge0\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow b^2+1\ge2b\) (3)

từ (1), (2) và (3) suy ra;

\(a^2+b^2+a^2+1+b^2+1\ge2ab+2a+2b\)

<=> \(2a^2+2b^2+2\ge2ab+2a+2b\)

<=> \(a^2+b^2+1\ge ab+a+b\) \(\xrightarrow[]{}\) đpcm

b, Ta xét;

\(a^2+b^2+c^2+d^2+e^2-ab-ac-ad-ae\)

\(=\left(\dfrac{a^2}{4}-ab+b^2\right)+\left(\dfrac{a^2}{4}-ac+c^2\right)+\left(\dfrac{a^2}{4}-ad+d^2\right)+\left(\dfrac{a^2}{4}-ae+e^2\right)\)

\(=\left(\dfrac{a}{2}-b\right)^2+\left(\dfrac{a}{2}-c\right)^2+\left(\dfrac{a}{2}-d\right)^2+\left(\dfrac{a}{2}-e\right)^2\)

mà \(\left\{{}\begin{matrix}\left(\dfrac{a}{2}-b\right)^2\ge0\\\left(\dfrac{a}{2}-c\right)^2\ge0\\\left(\dfrac{a}{2}-d\right)^2\ge0\\\left(\dfrac{a}{2}-e\right)^2\ge0\end{matrix}\right.\) => \(\left(\dfrac{a}{2}-b\right)^2+\left(\dfrac{a}{2}-c\right)^2+\left(\dfrac{a}{2}-d\right)^2+\left(\dfrac{a}{2}-e\right)^2\ge0\)

=> \(a^2+b^2+c^2+d^2+e^2-ab-ac-ad-ae\ge0\)

<=> \(a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\)

<=> \(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\) \(\xrightarrow[]{}\) đpcm

Answer 4

3, ĐKXĐ: a, b, c \(\ne0\) ; \(a+b+c\ne0\)

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

<=> \(\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)

<=> \(\left(a+b+c\right)\left(bc+ac+ab\right)=abc\)

<=> \(abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+abc+ac^2=abc\)

<=> \(a^2c+a^2b+b^2c+ab^2+bc^2+ac^2+2abc=0\)

<=>\(\left(a^2c+abc\right)+\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(bc^2+ac^2\right)=0\)

<=> \(ac\left(a+b\right)+ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)

<=> \(\left(a+b\right)\left(ac+ab+bc+c^2\right)=0\) <=> \(\left(a+b\right)\left[a\left(c+b\right)+c\left(b+c\right)\right]=0\)

<=> \(\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

Lại có: \(A=\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\left(b+c\right)\left(b^2-bc+c^2\right)\left(c+a\right)\left(c^2-ac+a^2\right)=0\) (Do \(\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\))

Vậy \(A=\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)=0\) khi \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)