M\(=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2bc}{b\left(ac+2c+2\right)}\)

M = \(\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{abc+2bc+2b}\)

M=\(\dfrac{1}{b+1+bc}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{2+2bc+2b}\)

M = \(\dfrac{1+b}{b+1+bc}+\dfrac{2bc}{2\left(1+bc+b\right)}\)

M = \(\dfrac{1+b}{b+1+bc}+\dfrac{bc}{b+1+bc}=\dfrac{1+b+bc}{b+1+bc}=1\)

nhân cả vế với abc ta có điều cần chứng minh

\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\ge\dfrac{ab+bc+ac}{2}\)

VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)

=>(đpcm)

mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\) 

chỉ cần chứng minh được \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé

nhân cả vế với abc ta có điều cần chứng minh

\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\ge\dfrac{ab+bc+ac}{2}\)

VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)

=>(đpcm)

mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)}{c\left(a+b\right)}\ge\dfrac{ab+bc+ac}{2}\)

chỉ cần chứng minh được\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+x}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé.

Thay abc = 2 vào biểu thức A ta được:

\(A=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc\cdot c}{ac+abc+abc}\\ A=\dfrac{1}{b+1+bc}+\dfrac{b}{bc+b+1}+\dfrac{bc}{1+bc+b}\\ A=\dfrac{1+b+bc}{1+b+bc}\\ A=1\)

\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)

\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)

\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)

b.

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)

\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)

ta có : \(A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(=\dfrac{a}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{abc^2+abc+ac}\)

\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)

\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Với a, b, c > 0 có:

\(P=\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\\ =\dfrac{a^2}{a\left(b+2c\right)}+\dfrac{b^2}{b\left(c+2a\right)}+\dfrac{c^2}{c\left(a+2b\right)}\)

\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\)

chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\) 

\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)

Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)

Cộng vế với vế:

\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)

Hint: \(M=3-Σ\dfrac{a^2+a+1}{a^2+2a+1}\)

dự đoán dấu đẳng thức xảy ra tại \(a=b=c=\sqrt{3}-1\) nên \(MinM=\dfrac{3\sqrt{3}}{4}\)ta đi chứng minh \(M\le\dfrac{3\sqrt{3}}{4}\)

Đặt \(a+1=x\); \(b+1=y\); \(c+1=z\)

\(\Rightarrow x+y+z=a+b+c+3=a+b+c+ab+bc+ac+abc+1\)

\(=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)\(=xyz\)

\(\Rightarrow x+y+z=xyz\) đặt \(x=tan\dfrac{A}{2};y=tan\dfrac{B}{2};z=tan\dfrac{C}{2}\)

\(\Rightarrow tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}=tan\dfrac{A}{2}.tan\dfrac{B}{2}.tan\dfrac{C}{2}\)(đúng , tự cm hoặc google)

do \(x=tan\dfrac{A}{2}\Rightarrow sinA=\dfrac{2x}{x^2+1}\Rightarrow\dfrac{sinA}{2}=\dfrac{x}{x^2+1}\)

\(\Rightarrow M=\Sigma\dfrac{sinA}{2}\)ta phai cm \(M=\Sigma\dfrac{sinA}{2}\le\dfrac{3\sqrt{3}}{4}\Leftrightarrow M\le\dfrac{3\sqrt{3}}{2}\)

\(\Leftrightarrow sinA+sinB+sinC\le\dfrac{3\sqrt{3}}{2}\)theo BDT Cauchy-Swarch

\(\Rightarrow sinA+sinB+sinC\le\sqrt{3\left(sinA^2+sin^2B+sin^2C\right)}\)

mặt khác \(sin^2A+sin^2B+sin^2C\le\dfrac{9}{4}\)(trong sach toan 10 co BDT nay hoac google)

\(\Rightarrow sinA+sinB+sinC\le\sqrt{\dfrac{3.9}{4}}=\dfrac{3\sqrt{3}}{2}\)

\(\Rightarrow dpcm\)

Chị cx học Tê Tiêu ạ,A mấy ạ