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

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

Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)

Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)

Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)

Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*

\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{​​}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)

\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)

Đẳng thức xảy ra khi a = b = c

P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:

1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)

\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)

bài 2 xem có ghi nhầm ko

Đề phải thêm là \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) nhé.

Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}.\)

\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a+b+c}{b+c+d}.\frac{a+b+c}{b+c+d}.\frac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\)

\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}.\)

\(\Rightarrow\frac{a}{d}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(đpcm\right).\)

Chúc bạn học tốt!

Nhanh giùm với ạ TTTT

mình làm rồi nhưng ngại trả lời ghê

Mình chỉ làm bài 1a, và bài 3 thôi nhé,còn lại là bạn tự làm nhé

Bài 1:

a, Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\Rightarrow\left[\frac{a}{b}\right]^2=\left[\frac{c}{d}\right]^2=\left[\frac{a+c}{b+d}\right]^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{(a+c)^2}{(b+d)^2}\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)

Bài 3 : Sửa đề : Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

CM : a = b = c

Cách 1 : Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

vì \(a+b+c\ne0\)

\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c\)

Do đó : \(a=b=c\).

Cách 2 : Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=m\), ta có : \(a=bm,b=cm,c=am\)

Do đó : \(a=bm=m(mc)=m\left[m(ma)\right]\)

\(\Rightarrow a=m^3a\Rightarrow m^3=1(a\ne0)\Rightarrow m=1\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

Cách 3 : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=\left[\frac{a}{b}\right]^3\Rightarrow1=\left[\frac{a}{b}\right]^3\Rightarrow\frac{a}{b}=1\)

Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^{2007}}{c^{2007}}=\frac{b^{2007}}{d^{2007}}=\frac{\left(a-b\right)^{2007}}{\left(c-d\right)^{2007}}.\)

mà \(\frac{a^{2007}}{c^{2007}}=\frac{b^{2007}}{d^{2007}}=\frac{a^{2007}+b^{2007}}{c^{2007}+d^{2007}}\)

=> đpcm

\(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\)\(\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\)\(\frac{a^{2007}}{c^{2007}}=\)\(\frac{b^{2007}}{c^{2007}}\)

\(\Rightarrow\)\(\frac{a^{2007}-b^{2007}}{c^{2007}-d^{2007}}=\frac{a^{2007}+c^{2007}}{c^{2007}+d^{2007}}\)

\(\Rightarrow\)\(\frac{\left(a-b\right)^{2007}}{\left(c-d\right)^{2007}}=\frac{a^{2007}+b^{2007}}{c^{2007}+d^{2007}}\)\((đpcm)\)

Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

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

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)