a) Áp dụng BĐT Cauchy-Schwarz dạng Engel: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Tương tự:\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c};\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)

Cộng theo vế 3 BĐT trên rồi chia cho 2 ta thu được đpcm

Đẳng thức xảy ra khi \(a=b=c\)

b)Đặt \(a+b=x;b+c=y;c+a=z\). Cần chứng minh:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

Cách làm tương tự câu a.

c) \(VT=\Sigma_{cyc}\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\Sigma_{cyc}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\le\frac{1}{16}\Sigma\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)

Đẳng thức xảy ra khi \(a=b=c=\frac{3}{4}\)

d) Em làm biếng quá anh làm nốt đi:P

Câu d : \(p=\frac{a+b+c}{2}\Rightarrow2p=a+b+c\)

Ta có : \(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\)

Tương tự : \(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\)

\(\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{b}\)

\(\Leftrightarrow2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

\(\Leftrightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\left(đpcm\right)\)

Dấu \("="\) xảy ra khi nó là tam giác đều

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, 

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

\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

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

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

+) Xét \(a+b+c+d=0\)

Suy ra : 

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

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

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

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

Do đó : \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{c+b}\)

\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}\)

\(M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(M=-4\)

+) Xét \(a+b+c+d\ne0\)

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

\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=4\)

Do đó : 

\(\frac{a+b+c+d}{a}=4\)\(\Leftrightarrow\)\(a+b+c+d=4a\) \(\left(1\right)\)

\(\frac{a+b+c+d}{b}=4\)\(\Leftrightarrow\)\(a+b+c+d=4b\) \(\left(2\right)\)

\(\frac{a+b+c+d}{c}=4\)\(\Leftrightarrow\)\(a+b+c+d=4c\) \(\left(3\right)\)

\(\frac{a+b+c+d}{d}=4\)\(\Leftrightarrow\)\(a+b+c+d=4d\) \(\left(4\right)\)

Từ (1), (2), (3) và (4) suy ra \(4a=4b=4c=4d\) \(\left(=a+b+c+d\right)\)

\(\Leftrightarrow\)\(a=b=c=d\)

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

\(\Rightarrow\)\(M=1+1+1+1=4\)

Vậy \(M=-4\) hoặc \(M=4\)

Chúc bạn học tốt ~ 

Ta có : 

\(2a+2b+2c=by+cz+ax+cz+ax+by\)

\(\Leftrightarrow\)\(2\left(a+b+c\right)=2\left(ax+by+cz\right)\)

\(\Leftrightarrow\)\(a+b+c=ax+by+cz\)

+) \(a+b+c=ax+\left(by+cz\right)=ax+2a=a\left(x+2\right)\)

\(\Rightarrow\)\(\frac{1}{x+2}=\frac{a}{a+b+c}\) \(\left(1\right)\)

+) \(a+b+c=by+\left(ax+cz\right)=by+2b=b\left(y+2\right)\)

\(\Rightarrow\)\(\frac{1}{y+2}=\frac{b}{a+b+c}\) \(\left(2\right)\)

+) \(a+b+c=cz+\left(ax+by\right)=cz+2c=c\left(z+2\right)\)

\(\Rightarrow\)\(\frac{1}{z+2}=\frac{c}{a+b+c}\) \(\left(3\right)\)

Từ (1), (2) và (3) suy ra \(M=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)

\(M=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)

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

Vậy \(M=1\)

Chúc bạn học tốt ~ 

a/ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ; \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\) ; \(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)

Cộng theo vế :

\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

b/ \(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\)

\(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{b+2c+a}\)

\(\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{4}{c+b+2a}\)

Cộng theo vế :

\(2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge4\left(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\right)\)

\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge2\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)