Áp dụng BĐT \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\ge0\)

\(\Leftrightarrow9abc+18\left(a+b+c\right)\ge12\left(ab+bc+ca\right)+27\)

\(\Leftrightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

Do đó:

\(P=4a^2+4b^2+4c^2+abc\ge4a^2+4b^2+4c^2+\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{2}{3}\left(a+b+c\right)^2+\dfrac{10}{3}\left(a^2+b^2+c^2\right)-3\)

\(P\ge\dfrac{2}{3}\left(a+b+c\right)^2+\dfrac{10}{9}\left(a+b+c\right)^2-3=13\)

Đề bài bạn viết thiếu số 1 bên vế phải rồi

Lời giải:

Áp dụng BĐT Schur:

$abc\geq (a+b-c)(b+c-a)(c+a-b)=(3-2a)(3-2b)(3-2c)$

$\Leftrightarrow 9abc\geq 12(ab+bc+ac)-27$

$\Leftrightarrow abc\geq \frac{4}{3}(ab+bc+ac)-3$

Do đó:

$4(a^2+b^2+c^2)+abc\geq 4(a^2+b^2+c^2)+\frac{4}{3}(ab+bc+ac)-3$

$=\frac{10}{3}(a^2+b^2+c^2)+\frac{2}{3}(a+b+c)^2-3$

$\geq \frac{10}{9}(a+b+c)^2+\frac{2}{3}(a+b+c)^2-3=13$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$

Bài đúng phải là $4a^2+4b^2+4c^2+abc\geq 13$ nhé bạn.

\(\Leftrightarrow\left(a+c\right)^2+b^2+2b\left(a+c\right)+\left(a+c\right)^2+b^2-2b\left(a+c\right)>4b^2\)

\(\Leftrightarrow\left(a+c\right)^2>b^2\)

\(\Leftrightarrow a+c>b\) (luôn đúng theo BĐT tam giác)

Vậy BĐT đã cho được chứng minh

Ta có:

Vế trái: -a.(c-d)-d.(a+c)

=-ac+ad-ad-cd

=-ac-cd (1)

Vế phải: -c(a+d)=-ac-cd (1)

Vì (1)=(2)

<=> -a.(c-d)-d.(a+c)=-c.(a+d) (đpcm)

(Lưu ý: "đpcm" nghĩa là "điều phải chứng minh".)

Lời giải:

1) \(VT=-a.\left(c-d\right)-d.\left(a+c\right)\)

$=-ac+ad-da-dc$

$=-ac-dc$

$=-c(a+d) (đpcm)$

$2) (3a+2).(2a-1)+(3-a).(6a+2)-17.(a-1)$

$=6a^2-3a+4a-2+18a+6-6a^2-2a-17a+17$

$=21$

Vậy giá trị biểu thức không phụ thuộc vào a

Ta có : $(3a+2)(2a-1)+(3-a)(6a+2)-17(a-1)$

$=6a^2+a-2+18a+6-6a^2-2a-17a+17$

$=21$ không phụ thuộc vào a.

\(B=\dfrac{\left(4a^2-1\right)\left(b-c\right)-\left(4b^2-1\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\dfrac{4c^2-1}{\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{4a^2b-4a^2c-b+c-4ab^2+4b^2c+a-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\dfrac{4ac^2-4bc^2-a+b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2b-4a^2c+a-b-4ab^2+4b^2c+4ac^2-4bc^2-a+b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2b-4ab^2-4a^2c+4ac^2-4bc^2+4b^2c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2\left(b-c\right)+4bc\left(b-c\right)-4a\left(b^2-c^2\right)}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2+4bc-4a\left(b+c\right)}{\left(a-c\right)\left(a-b\right)}\)

\(=\dfrac{4a^2-4ab+4bc-4ac}{\left(a-c\right)\left(a-b\right)}\)

\(=\dfrac{4a\left(a-b\right)-4c\left(a-b\right)}{\left(a-c\right)\left(a-b\right)}=4\)

1.

Ta có x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x.

\(\left(\frac{x}{y}+1\right)\left(\frac{y}{z}+1\right)\left(\frac{z}{x}+1\right)\)\(=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{z+x}{x}\)\(=-\frac{xyz}{xyz}=-1\)

2) a+b+c=0 <=> (a+b+c)^2=0

<=> a^2+b^2+c^2+2(ab+bc+ca)=0

VT >= ab+bc+ca+2(ab+bc+ca)

=> 0 >= 3(ab+bc+ca)

<=> 0 >= (ab+bc+ca) 

Dấu "=" xảy ra khi a=b=c=0