\(a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\ge0.\)

\(a^2\left(\frac{a-}{b+c}\frac{b}{a+c}\right)+b^2\left(\frac{b}{a+c}\frac{-c}{a+b}\right)+c^2\left(\frac{c-}{a+b}\frac{a}{b+c}\right)\ge0.\)

\(a^2\left(a^2-b^2\right)+b^2\left(b^2-c^2\right)+c^2\left(c^2-a^2\right)\ge0.\)

\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2.\) cái này dễ rồi .

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

áp dụng dbt cosi cho 2 số:\(\frac{a^3}{b^2}\)va b ta duoc :

\(\frac{a^3}{b^2}\)+a\(\ge\)2\(\sqrt{\frac{a^3}{b^2}.a}\)=2\(\frac{a^2}{b}\)

CMTT:\(\frac{b^3}{c^2}\)+b\(\ge\)2\(\frac{b^2}{c}\)

\(\frac{c^3}{a^2}\)+c\(\ge\)2\(\frac{c^2}{a}\)

\(\Rightarrow\)\(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{c^3}{a^2}\)+(a+b+c)\(\ge\)2(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\))

\(\Leftrightarrow\)\(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{c^3}{a^2}\)\(\ge\)2(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)) - (a+b+c)           (1)

Ap dụng bdt cosi cho các số dương , ta được:

\(\frac{a^2}{b}\)+\(b\)\(\ge\)2\(\sqrt{\frac{a^2}{b}.b}\)=2a

CMTT: \(\frac{b^2}{c}\)+c\(\ge\)2b

\(\frac{c^2}{a}\)+a\(\ge\)2c

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

\(\Leftrightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)\(\ge\)a+b+c 

\(\Leftrightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\) _ (a + b + c )  \(\ge\)0

Do Đó:TỪ (1) ta co : \(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{b^3}{c^2}\)\(\ge\)(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\) )

Xét hiệu hai vế:

BĐT \(\Leftrightarrow\left(\frac{a^3}{b^2}-\frac{a^2b}{b^2}\right)+\left(\frac{b^3}{c^2}-\frac{b^2c}{c^2}\right)+\left(\frac{c^3}{a^2}-\frac{c^2a}{a^2}\right)-\left(a+b+c-b-c-a\right)\ge0\)

\(\Leftrightarrow\left(\frac{a^3}{b^2}-\frac{a^2b}{b^2}\right)+\left(\frac{b^3}{c^2}-\frac{b^2c}{c^2}\right)+\left(\frac{c^3}{a^2}-\frac{c^2a}{a^2}\right)-\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]\ge0\)

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

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

BĐT này đúng với mọi a,b,c > 0 nên ta có Q.E.D

Dấu "=" xảy ra khi a =b =c

P/s: Toán 7 gì mà khó thế nhỉ??Mình cũng không chắc đâu nha!

trả lời 

=>  bé hơn hoặc bằng 0 

chúc bn

thành công trong 

học tập

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2b^2}{b^2c^2}}=2\left|\frac{a}{c}\right|\ge\frac{2a}{c}\)

Tương tự: \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{2c}{b}\) ; \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)

Cộng vế với vế:

\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

mình nghĩ đề bài sai một chỗ :\(\frac{a^2}{b^2}\)chứ ko phải là \(\frac{a}{b^2}\)

khó quá chưa học

mình chịu thôi bạn ơi mình mới học lớp 5 ak

a) \(\frac{a}{c}=\frac{c}{b}\)

\(\Rightarrow\frac{a^2}{c^2}=\frac{c^2}{b^2}=\frac{a^2+c^2}{c^2+b^2}=\frac{a}{c}.\frac{c}{b}=\frac{a}{b}\)

b) \(\frac{a}{c}=\frac{c}{b}\)\(\Rightarrow ab=c^2\)

\(\frac{b^2-a^2}{a^2+c^2}=\frac{b^2-ab+ab-a^2}{a^2+ab}=\frac{\left(b-a\right)b+\left(b-a\right)a}{a.\left(a+b\right)}=\frac{\left(b-a\right)\left(b+a\right)}{a.\left(a+b\right)}=\frac{b-a}{a}\)

cho mk hỏi viết phan số bằng cách nào vậy