\(Q=\frac{c+ab}{a+b}+...+\frac{b+ac}{a+c};\frac{c+ab}{a+b}=\frac{ca+cb+c^2+ab}{a+b}=\frac{\left(c+b\right)\left(c+a\right)}{a+b}\)

\(\text{tương tự ta có:}2Q=\frac{2\left(a+b\right)\left(b+c\right)}{a+c}+\frac{2\left(b+c\right)\left(a+c\right)}{a+b}+\frac{2\left(a+b\right)\left(a+c\right)}{b+c}\)

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

\(=2\left[2\left(a+b+c\right)\right]=4\Rightarrowđpcm\text{ dấu "=":}a=b=c=\frac{1}{3}\)

\(pt\Leftrightarrow x^2-4=x^2-2xm+m^2\Leftrightarrow2xm-m^2=4\Leftrightarrow2xm-4=m^2\)

m-4/m=2x

nhầm tí

\(m=x-\sqrt{x^2-4}\)

với x=<-2;x>=2

Theo giả thiết, ta có: \(ab+bc+ca+abc=4\)\(\Leftrightarrow\left(ab+bc+ca\right)+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8\)\(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)

Áp dụng bất đẳng thức Cauchy, ta được: \(P=\frac{4}{\left[\left(a+b\right)^2+4\right]+16}+\frac{4}{\left[\left(b+c\right)^2+4\right]+16}+\frac{4}{\left[\left(c+a\right)^2+4\right]+16}\)\(\le\frac{4}{4\left(a+b\right)+16}+\frac{4}{4\left(b+c\right)+16}+\frac{4}{4\left(c+a\right)+16}\)\(=\frac{1}{\left(a+2\right)+\left(b+2\right)}+\frac{1}{\left(b+2\right)+\left(c+2\right)}+\frac{1}{\left(c+2\right)+\left(a+2\right)}\)\(\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)

Đẳng thức xảy ra khi a = b = c = 1

BT1:

\(1,2FeS_2+\frac{11}{2}O_2\underrightarrow{t}Fe_2O_3+4SO_2\)

\(2,2SO_2+O_{_{ }2}\underrightarrow{t,V_2O_5}2SO_3\)

\(3,SO_3+H_2O\rightarrow H_2SO_4\)

\(4,H_2SO_4+Na_2SO_3\rightarrow Na_2SO_4+SO_2+H_2O\)

\(5,CaO+SO_2\underrightarrow{t}CaSO_3\)

\(6,CaSO_3\underrightarrow{t,xt}CaO+SO_2\)

\(7,CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(8,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Vẽ đường phân giác AD, gọi H là chân đường vuông góc kẻ từ B xuống AD

Theo tính chất đường phân giác, ta có: \(\frac{AB}{BD}=\frac{AC}{CD}=\frac{AB+AC}{BC}=\frac{b+c}{a}\Rightarrow\frac{BD}{AB}=\frac{a}{b+c}\)

Suy ra \(\sin\frac{A}{2}=\sin BAD=\frac{BH}{BA}\le\frac{BD}{AB}=\frac{a}{b+c}\)(đpcm)

Kiệt nguyễn

\(\frac{sinA}{2}\ne sin\frac{A}{2}\)