Cauchy Schwars 

\(M\ge\frac{\left(1+1+1\right)^2}{\left(a+b+c\right)^2}=\frac{9}{\left(a+b+c\right)^2}\ge9\Rightarrow M_{min}=9\Leftrightarrow a=b=c=\frac{1}{3}\)

\(M=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{\left(a+b+c\right)^2}\ge9\)

Dau '=' xay ra khi \(a=b=c=\frac{1}{3}\)

Vay \(M_{min}=9\)

a,b,c khác nhau đôi một nghĩa là từng cặp số khác nhau ,là:

+a khác b

+b khác c

+c khác a

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

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0\)

Suy ra: \(ab==-\left(bc+ac\right)=-bc-ac\)

    \(bc=-\left(ab+ac\right)=-ab-ac\)

\(ac=-\left(ab+bc\right)=-ab-bc\)

Nên \(a^2+2ab=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự,ta cũng có: \(b^2+2ac=\left(b-a\right)\left(b-c\right)\)

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

Vậy \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

những câu còn lại tương tự,bn tự làm nhé
 

Nhân cả 2 vế với a+b+c 

Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)

\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0

dễ rồi nhé

b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được: 

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)

=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)

=>P_max=3/4 <=> x=y=z=1/3

Ta có: \(5a^2+2ab+2b^2=4a^2+2ab+b^2+\left(a^2+b^2\right)\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)

Lại có: \(\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)

Tương tự cộng lại ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Theo BĐT Bunhiacopxki ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{3}\)

\(\Rightarrow VT\le\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\)

Dấu = xảy ra khi \(a=b=c=\sqrt{3}\)

khó quá xin lỗi nha em  mới hok lớp 7

Câu này lớp 7 tớ có làm. Cũng như cái mà gọi là áp dụng t/c dãy tỉ số bằng nhau và tỉ lệ thức. mình tính ra dc a, b. c rồi.

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

\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)

\(\Leftrightarrow bc+ca+ab=0\)

\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)

Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)

\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)

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

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

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

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

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

\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

a, b, c chưa khác 0 bạn nhé

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

\(\Leftrightarrow abc.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Leftrightarrow\hept{\begin{cases}bc=-\left(ab+ac\right)\\ab=-\left(bc+ac\right)\\ac=-\left(bc+ab\right)\end{cases}}\)

Ta có: \(a^2+2bc=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(c-a\right)\left(c-b\right)\)

\(\Leftrightarrow N=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)

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