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

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

Bài 1: \(a+b\ge1\). cm \(a^4+b^4\ge\dfrac{1}{8}\)

ta có : \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)(BĐT bunyakovsky)

Áp dụng BĐt bunyakovsky 1 lần nữa:

\(a^4+b^4\ge\dfrac{1}{2}\left(a^2+b^2\right)^2\ge\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}\)

dấu = xảy ra khi \(a=b=\dfrac{1}{2}\)

Bài 2:

Áp dụng BĐT bunyakovsky dạng đa thức và phân thức:

\(\left(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\right)\left(a+b+c\right)\ge\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)^2\ge\left[\dfrac{\left(a+b+c\right)^2}{a+b+c}\right]^2=\left(a+b+c\right)^2\)

do đó \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\)

dấu = xảy ra khi a=b=c

Bài 1:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)

Lại theo Cauchy-Schwarz lần nữa:

\(\left[\left(1^2\right)^2+\left(1^2\right)^2\right]\left[\left(a^2\right)^2+\left(b^2\right)^2\right]\ge\left(a^2+b^2\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow2\left(a^4+b^4\right)\ge\dfrac{1}{4}\Leftrightarrow a^4+b^4\ge\dfrac{1}{8}\)

Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)

Bài 2:

Trước tiên ta chứng minh \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\)

Ta chứng minh bổ đề: \(\dfrac{a^3}{b^2}\ge\dfrac{a^2}{b}+a-b\)

\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng)

Viết các BĐT tương tự và cộng lại

\(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge\dfrac{a^2}{b}+a-b+\dfrac{b^2}{c}+b-c+\dfrac{c^2}{a}+c-a=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\left(1\right)\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

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

Từ \((1);(2)\) ta thu được ĐPCM

C1:Áp dụng Bất đẳng thức AM-GM ta có:

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1^2}{a+b}+\dfrac{1^2}{b+c}+\dfrac{1^2}{c+a}\ge\)

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

\(\Rightarrow A=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\left(a+b+c\right).\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9}{2}\)Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

C2: Khai triển

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

\(=1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}\) (bn tự khai triển đầy đủ nha)

Áp dụng BĐT Nesbitt ta có:

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

\(\left(1+1+1\right)+\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=9^2\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Rightarrow a^2+b^2+c^2\ge3\)

Lại có: \(a^2+b^2+c^2\ge ab+bc+ac\forall a,b,c\)

\(\Rightarrow3\ge ab+bc+ac\Rightarrow ab+bc+ac\le3\)

Bất đẳng thức ban đầu tương đương với:

\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

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

Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}a\left(b^2+1\right)\ge a\cdot2\sqrt{b^2}=2ba\\b\left(c^2+1\right)\ge b\cdot2\sqrt{c^2}=2cb\\c\left(a^2+1\right)\ge c\cdot2\sqrt{a^2}=2ac\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

Mà \(ab+bc+ca\le3\)\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\cdot3}=\dfrac{9}{6}=\dfrac{3}{2}\)

Đẳng thức xảy ra khi \(a=b=c=1\)

\(VT=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\)

\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)

\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

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

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

\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)

\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) (1)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow3\ge ab+bc+ca\)

\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)

\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\)(2)

Từ (1) và (2)

\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )

Dấu "=" xảy ra khi \(a=b=c=1\)

cần giúp ko

cách khác

Chứng minh điều ngược lại

Câu 2: phần (b)

\(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{\left(a+b\right)^2}{ab}\)

Để tồn tại A => a,b khác 0

\(\left(a+b\right)^2\ge0\forall a,b\in R\)

\(ab>0\) có một chút hy vọng \(A\ge4\)

ab< 0 tắt hét hy vọng \(A\ge4\)

Kết luận chẳng hy vọng gì

cần face với thực tế Quý Minh mình đã giải thích cho bạn thế nào là face đó

Bác google được sinh ra để làm gì, đăng nhiều vc, google có hết mà ;v

Bài 1,2,3,4 đơn giản, tự làm :v

7) \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=\dfrac{abc}{c^3}+\dfrac{abc}{a^3}+\dfrac{abc}{b^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{1}{3abc}=\dfrac{1}{3}\)

P/S: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

5) ĐK: a>b>0

\(3a^2+3b^2=10ab\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)

Tự phân tích

Mà a>b>0=> Chọn a=3b

Thay vào

Bài 6 tương tự bài 5

Có bất mãn chỗ nào thì ib nha bạn :))