a)

\(a^2+b^2+c^2+d^2+m^2-a(b+c+d+m)\)

\(=\frac{4a^2+4b^2+4c^2+4d^2+4m^2-4a(b+c+d+m)}{4}\)

\(=\frac{(a^2+4b^2-4ab)+(a^2+4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4m^2-4am)}{4}\)

\(=\frac{(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2m)^2}{4}\geq 0\) (đpcm)

Dấu "=" xảy ra khi \(a=2b=2c=2d=2m\)

b)

Xét hiệu

\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}=\frac{x+y}{xy}-\frac{4}{x+y}=\frac{(x+y)^2-4xy}{xy(x+y)}\)

\(=\frac{x^2+y^2-2xy}{xy(x+y)}=\frac{(x-y)^2}{xy(x+y)}\geq 0, \forall x,y>0\)

\(\Rightarrow \frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\) (đpcm)

Dấu "=" xảy ra khi $x=y$

c)

Xét hiệu:

\((a^2+c^2)(b^2+d^2)-(ab+cd)^2\)

\(=(a^2b^2+a^2d^2+c^2b^2+c^2d^2)-(a^2b^2+2abcd+c^2d^2)\)

\(=a^2d^2-2abcd+b^2c^2=(ad-bc)^2\geq 0\)

\(\Rightarrow (a^2+c^2)(b^2+d^2)\geq (ab+cd)^2\) (đpcm)

Dấu "=" xảy ra khi \(ad=bc\)

d)

Xét hiệu:

\(a^2+b^2-(a+b-\frac{1}{2})=a^2+b^2-a-b+\frac{1}{2}\)

\(=(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})\)

\(=(a-\frac{1}{2})^2+(b-\frac{1}{2})^2\geq 0\)

\(\Rightarrow a^2+b^2\geq a+b-\frac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

Ùi mình làm theo kiểu khác thử :V, nhưng có hơi hướng giống và bổ sung :D

Câu 2 : a,b,c > 0. CM : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Giải :

C1 : Áp dụng bất đẳng thức Cauchy - Schwarz dạng Engel ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\left(ĐPCM\right)\)

Đẳng thức xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\).

C2 : Đầy đủ hơn với cách giải đúng của bạn Hoàng Thiên Di :

Áp dụng BĐT AM-GM cho 3 số dương (sgk là cosi :v)

\(a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

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

\(\ge3+2+2+2=9\left(ĐPCM\right)\)

Câu 3 : a,b,c > 0. CM : \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

Giải :

\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

\(\Leftrightarrow\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\ge6\)

\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge6\)

Theo bất đẳng thức Cosi : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{yx}}=2\)

Thay vào các vế được : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\sqrt{1}=2\)

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\sqrt{1}=2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\sqrt{1}=2\)

\(\Leftrightarrow2+2+2\ge6\) (đúng)

BĐT được c/m.

xem lại đề

a=b=c=1 =>3<=2

a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)

\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1

b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)

\(\Leftrightarrow x< 5\)

c) \(\dfrac{-1}{2x+3}< 0\)

dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp

2a)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)

\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)

Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)

\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )

Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)

\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

2b)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=1\)

Bài 1)

Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Giải

Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0

Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)

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

\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)

Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y

a.

\(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2x+10-x^2-5x=0\)

\(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

b.

\(2x^2+3x-5=0\)

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)

bài 2:

ĐKXĐ: x khác -1

\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)

\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow0x=-1\)

\(\Leftrightarrow x\in\varnothing\)

Suy ra pt vô nghiệm

b.

ĐKXĐ: x khác \(\dfrac{3}{2}\)

\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)

\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)

5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)

áp dụng bđ cosy

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

=> đpcm

6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

hay với mọi x thuộc R đều là nghiệm của bpt

7.áp dụng bđt cosy

\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)

1. (a-b)²>=0

=> a²+b²-2ab>=0

2. (a-b)²>=0

=> a²+b²>=2ab

=> \(\dfrac{a^2 +b^2}{2}\ge ab\)

3.Ta phích ra thôi,ta được : a²+2a < a²+2a+1

=> cauis trên đúng

1) 2( a² + b² ) ≥ ( a + b)²

<=> 2a² + 2b² - a² - 2ab - b² ≥ 0

<=> a² - 2ab + b² ≥ 0

<=> ( a - b )² ≥ 0 ( luôn đúng )

=> đpcm

2) Áp dụng BĐT Cô-si cho 2 số dương x , y , ta có :

a + b ≥ \(2\sqrt{ab}\)

=> \(\dfrac{1}{x}+\dfrac{1}{y}\) ≥ 2\(\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\)

=> ( x + y)( \(\dfrac{1}{x}+\dfrac{1}{y}\) ) ≥ \(2\sqrt{xy}\)2\(\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\)

=> ( x + y)( \(\dfrac{1}{x}+\dfrac{1}{y}\)) ≥ 4

=> \(\dfrac{1}{x}+\dfrac{1}{y}\) ≥ \(\dfrac{4}{x+y}\)

Bài 3:

a) Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\) \(\geq 2.\frac{(1+1)^2}{2xy+x^2+y^2}=\frac{8}{(x+y)^2}=8\)

Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)

b) Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{2xy}+\left (\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\geq \frac{1}{2xy}+\frac{(1+1)^2}{2xy+x^2+y^2}\)

\(=\frac{1}{2xy}+\frac{4}{(x+y)^2}\)

Theo BĐT AM-GM:

\(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{1}{2xy}\geq 2\)

Do đó \(\frac{1}{xy}+\frac{1}{x^2+y^2}\geq 2+4=6\)

Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)

Bài 1: Thiếu đề.

Bài 2: Sai đề, thử với \(x=\frac{1}{6}\)

Bài 4 a) Sai đề với \(x<0\)

b) Áp dụng BĐT AM-GM:

\(x^4-x+\frac{1}{2}=\left (x^4+\frac{1}{4}\right)-x+\frac{1}{4}\geq x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2\geq 0\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x^4=\frac{1}{4}\\ x=\frac{1}{2}\end{matrix}\right.\) (vô lý)

Do đó dấu bằng không xảy ra , nên \(x^4-x+\frac{1}{2}>0\)

Bài 6: Áp dụng BĐT AM-GM cho $6$ số:

\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^3b^3c^3d^3}=6\)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c=d=1\)

5) a) Đặt b+c-a=x;a+c-b=y;a+b-c=z thì 2a=y+z;2b=x+z;2c=x+y

Ta có:

\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)

Vậy ta suy ra đpcm

b) Ta có: a+b>c;b+c>a;a+c>b

Xét: \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)

.Tương tự:

\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c};\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)

Vậy ta có đpcm

6) Ta có:

\(a^2+b^2+c^2+d^2+ab+cd\ge2ab+2cd+ab+cd=3\left(ab+cd\right)\)

\(ab+cd=ab+\dfrac{1}{ab}\ge2\)

Suy ra đpcm