trả lời

dùng bất đẳng thức cosi đc ko

hok tốt

undefined la gi

ta có

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

\(\Leftrightarrow a=b=c=>\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=1\)

tương tự 

\(\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\ge1\)

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

=>\(1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\ge2\)

=> dpcm

Không mất tính tổng quát, chuẩn hóa a + b + c = 1

Khi đó, ta cần chứng minh: \(\frac{\left(a+1\right)^2}{2a^2+\left(1-a\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(1-b\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(1-c\right)^2}\le8\)

Xét bất đẳng thức phụ: \(\frac{\left(x+1\right)^2}{2x^2+\left(1-x\right)^2}\le4x+\frac{4}{3}\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{\left(3x-1\right)^2\left(4x+1\right)}{2x^2+\left(1-x\right)^2}\ge0\)*đúng*

Áp dụng, ta được: \(\frac{\left(a+1\right)^2}{2a^2+\left(1-a\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(1-b\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(1-c\right)^2}\)\(\le4\left(a+b+c\right)+4=4.1+4=8\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi a = b = c

Chuẩn hóa ta có : \(a+b+c=3\)

=> \(\frac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}=\frac{\left(a+3\right)^2}{2a^2+\left(3-a\right)^2}=\frac{a^2+6a+9}{3\left(a^2-2a+3\right)}\)

Xét\(\frac{a^2+6a+9}{3\left(a^2-2a+3\right)}\le\frac{4}{3}a+\frac{4}{3}\)

<=> \(a^2+6a+9\le4\left(a+1\right)\left(a^2-2a+3\right)\)

<=> \(4a^3-5a^2-2a+3\ge0\)

<=> \(\left(a-1\right)^2\left(4a+3\right)\ge0\)luôn đúng

Khi đó 

\(VT\le\frac{4}{3}\left(a+b+c\right)+4=\frac{4}{3}.3+4=8\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

bài lớp 10 em chưa hok nha anh

\(\dfrac{b+c-a}{2a}+\dfrac{a-b+c}{2b}+\dfrac{a+b-c}{2c}\ge\dfrac{3}{2}\)

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

Viết lại BĐT cần chứng minh như sau:

\(\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}+\dfrac{a}{2b}-\dfrac{1}{2}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{1}{2}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{3}{2}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge0\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{b}{2a}+\dfrac{a}{2b}=\dfrac{1}{2}\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{a}\cdot\dfrac{a}{b}}=2\cdot\dfrac{1}{2}=1\)

\(\dfrac{c}{2a}+\dfrac{a}{2c}=\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{c}{a}+\dfrac{a}{c}}=\dfrac{1}{2}\cdot2=1\)

\(\dfrac{b}{2c}+\dfrac{c}{2b}=\dfrac{1}{2}\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{c}\cdot\dfrac{c}{b}}=\dfrac{1}{2}\cdot2=1\)

Cộng theo vế 3 BĐT trên ta có:

\(\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}\ge3\)

\(\Rightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge3-3=0\)

BĐT đúng nên ta có ĐPCM

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

bạn ơi a2 là a^2 bạn nhé,mấy cái khác cũng tương tự,vì mình lười bấm nhé)

A=2a2b2+2b2c2+2a2c2−a4−b4−c4

⟺A=4a2c2−(a4+b4+c4−2a2b2+2a2c2−2b2c2)

⟺A=4a2c2−(a2−b2+c2)2

⟺A=(2ac+a2−b2+c2)(2ac−a2+b2−c2)

⟺A=((a+c)2−b2)(b2−(a−c)2)

⟺A=(a+b+c)(a+c−b)(b+a−c)(b−a+c)

Mà a, b, ca, b, c là 33 cạnh của tam giác nên:a+b+c>0;a+c−b>0;b+a−c>0;b−a+c>0⟹(a+b+c)(a+c−b)(b+a−c)(b−a+c)>0
⟹A>0 (Dpcm)