hơn 1 năm rồi không ai làm :'(

a) Áp dụng bđt Cauchy ta có :

\(a+b\ge2\sqrt{ab}\)(1)

\(b+c\ge2\sqrt{bc}\)(2)

\(c+a\ge2\sqrt{ca}\)(3)

Nhân (1), (2), (3) theo vế

=> \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{a^2b^2c^2}=8\sqrt{\left(abc\right)^2}=8\left|abc\right|=8abc\)

=> đpcm

Dấu "=" xảy ra <=> a=b=c

b) Áp dụng bđt AM-GM ta có :

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2\sqrt{c^2}=2c\)

TT : \(\frac{ca}{b}+\frac{ab}{c}\ge2a\); \(\frac{bc}{a}+\frac{ab}{c}\ge2b\)

Cộng vế với vế

=> \(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)

=> \(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

a)Sắp xếp:a\(\ge\) b\(\ge\) c\(\ge\) 0

a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)

=a(a-b)[(a-b)=(b-c)]-b(a-b)(b-c)=c(a-c)(b-c)

=a(a-b)²+a(a-b)(b-c)-b(a-b)(b-c)+c(a-c)(b-c)

=a(a-b)²+(b-c)(a-b)²+c(a-c)(b-c)\(\ge\) 0

a) Giả sử:

\(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Rightarrow\frac{a^2+2ab+b^2}{4}\ge ab\)

\(\Rightarrow\frac{a^2+2ab+b^2}{4}-ab\ge0\)

\(\Rightarrow\frac{\left(a-b\right)^2}{4}\ge0\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng )

=> đpcm

b, Bất đẳng thức Cauchy cho các cặp số dương \(\frac{bc}{a}\)và \(\frac{ca}{b};\frac{bc}{a}\)và \(\frac{ab}{c};\frac{ca}{b}\)và \(\frac{ab}{c}\)

Ta lần lượt có : \(\frac{bc}{a}+\frac{ca}{b}\ge\sqrt[2]{\frac{bc}{a}.\frac{ca}{b}}=2c;\frac{bc}{a}+\frac{ab}{c}\ge\sqrt[2]{\frac{bc}{a}.\frac{ab}{c}}=2b;\frac{ca}{b}+\frac{ab}{c}\ge\sqrt[2]{\frac{ca}{b}.\frac{ab}{c}}\)

Cộng từng vế ta đc bất đẳng thức cần chứng minh . Dấu ''='' xảy ra khi \(a=b=c\)

c, Với các số dương \(3a\) và \(5b\), Theo bất đẳng thức Cauchy ta có \(\frac{3a+5b}{2}\ge\sqrt{3a.5b}\)

\(\Leftrightarrow\left(3a+5b\right)^2\ge4.15P\)( Vì \(P=a.b\)) 

\(\Leftrightarrow12^2\ge60P\)\(\Leftrightarrow P\le\frac{12}{5}\Rightarrow maxP=\frac{12}{5}\)

Dấu ''='' xảy ra khi \(3a=5b=12:2\)

\(\Leftrightarrow a=2;b=\frac{6}{5}\)

bạn ơi, bài này sai đề rồi

Ta có: BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow\dfrac{2a-\left(a+b\right)}{2\left(a+b\right)}+\dfrac{2b-\left(b+c\right)}{2\left(b+c\right)}+\dfrac{2c-\left(c+a\right)}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{\left(b+c\right)\left(c+a\right)}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\) (đúng)

Vậy BĐT luôn đúng với \(a\ge b\ge c>0\)

a/ \(\Leftrightarrow a^2-b^2+c^2\ge a^2+b^2+c^2-2ab+2ac-2bc\)

\(\Leftrightarrow b^2-ab+ac-bc\le0\)

\(\Leftrightarrow b\left(b-a\right)-c\left(b-a\right)\le0\)

\(\Leftrightarrow\left(b-c\right)\left(b-a\right)\le0\) (luôn đúng do \(a\ge b\ge c\))

Dấu "=" xảy ra khi \(\left[{}\begin{matrix}a=b\\b=c\end{matrix}\right.\)

b/ Tương tự như câu trên:

\(a^2-b^2+c^2-d^2\ge\left(a-b+c\right)^2-d^2=\left(a-b+c-d\right)\left(a-b+c+d\right)\ge\left(a-b+c-d\right)^2\)

a)\(\dfrac{\left(a+b\right)^2}{4}\ge ab\)\(\Leftrightarrow\dfrac{a^2+ab+b^2}{4}\ge0\)\(\Leftrightarrow\dfrac{\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}}{4}\ge0\left(đpcm\right)\)

Vậy \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

b) Áp dụng Cauchy, ta có:

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

Tương tự: \(\dfrac{ca}{b}+\dfrac{ab}{c}\ge2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)

Cộng vế theo vế các BĐT vừa chứng minh rồi rút gọn ta được đpcm.

c) ta có \(3a+5b=12\Rightarrow a=\dfrac{12-5b}{3}=4-\dfrac{5b}{3}\)

\(\Rightarrow P=ab=\left(4-\dfrac{5b}{3}\right)b=4b-\dfrac{5b^2}{3}\)

\(\Rightarrow15P=60b-25b^2=36-\left(25b^2-60b+36\right)=36-\left(5b-6\right)^2\)

\(\Rightarrow15P\le36\Rightarrow P\le\dfrac{36}{15}=\dfrac{12}{5}\) Vậy GTLN của \(P=\dfrac{12}{5}\) tại \(a=2;b=\dfrac{6}{5}\)

Ta có: \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{a}{2b}+\dfrac{b}{2c}+\dfrac{c}{2a}=\dfrac{1}{2}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\ge\dfrac{1}{2}.3=\dfrac{3}{2}\) ( BĐT AM - GM )

Dấu " = " khi a = b = c

\(\Rightarrowđpcm\)

BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{2}\cdot\dfrac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{\left(b+c\right)\left(c+a\right)}\right)\ge0\)

\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{\left(c-a\right)\left(c+a\right)+\left(a-c\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(luôn đúng)

\(\Rightarrowđpcm\)

tham khảo tại đây-_-

Câu hỏi của Nguyễn Thị Bình Yên - Toán lớp 8 | Học trực tuyến

Bài 1: Áp dụng BĐT Cauchy cho 3 số dương:

\(VT\ge3\sqrt[3]{\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{abc}}\ge3\sqrt[3]{\frac{8abc}{abc}}=6\) (đpcm)

Giải phần dấu "=" ra ta được a = b =c

Bài 2: Đặt \(a+b=x;b+c=y;c+a=z\)

Suy ra \(a=\frac{x-y+z}{2};b=\frac{x+y-z}{2};c=\frac{y+z-x}{2}\)

Suy ra cần chứng minh \(\frac{x-y+z}{2y}+\frac{x+y-z}{2z}+\frac{y+z-x}{2x}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{x+z}{2y}+\frac{x+y}{2z}+\frac{y+z}{2x}\ge3\)

\(\Leftrightarrow\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\ge6\)

Bài toán đúng theo kết quả câu 1.