Bài này dễ mà bạn

dễ thì bn giải hộ mk đi,nói đc lm đc nhỉ

cac cap tam giac co dien h bang nhau la AOB va BOC. Vi co cap song song voi nhau va cat toi diem O

bạn Phạm Thị Thúy Phượng gửi nhầm bài rồi 

\(a\left(2a-1\right)+b\left(2b-1\right)=2ab\)

\(\Leftrightarrow2a^2+2b^2-a-b=2ab\le\frac{\left(a+b\right)^2}{2}\)

Mà \(2a^2+2b^2\ge\left(a+b\right)^2\)

Đặt \(a+b=t\Rightarrow t^2-t\le\frac{t^2}{2}\Leftrightarrow t^2-t\le0\Leftrightarrow t\le1\Rightarrow a+b\le1\)

\(F=\frac{a^3+2020}{b}+\frac{b^3+2020}{a}=\frac{a^3}{b}+\frac{b^3}{a}+2020\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(=\frac{a^4+b^4}{ab}+2020\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{\left(a+b\right)^4}{2\left(a+b\right)^2}+\frac{8080}{a+b}\)

\(=\frac{\left(a+b\right)^2}{2}+\frac{8080}{a+b}=\frac{\left(a+b\right)^2}{2}+\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(a+b\right)}+\frac{8079}{a+b}\)

\(\ge3\sqrt[3]{\frac{\left(a+b\right)^2}{8\left(a+b\right)^2}}+\frac{8079}{1}=\)

đoạn cuối bí nhá 

GT => (a+1)(b+1)(c+1)=(a+1)+(b+1)+(c+1)

Đặt \(\frac{1}{a+1}=x,\frac{1}{1+b}=y,\frac{1}{c+1}=z\), ta cần tìm min của\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\)với xy+yz+zx=1

\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\Leftrightarrow\frac{2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Mà  (x+y)(y+z)(z+x) >= 8/9 (x+y+z)(xy+yz+xz) >= \(\frac{8\sqrt{3}}{9}\) nên \(M\)=< \(\frac{3\sqrt{3}}{4}\),dấu bằng xảy ra khi a=b=c=\(\sqrt{3}-1\)

Theo giả thiết, ta có: \(abc+ab+bc+ca=2\)

\(\Leftrightarrow abc+ab+bc+ca+a+b+c+1=a+b+c+3\)

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

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

Đặt \(\left(a+1;b+1;c+1\right)\rightarrow\left(\frac{\sqrt{3}}{x};\frac{\sqrt{3}}{y};\frac{\sqrt{3}}{z}\right)\). Khi đó giả thiết bài toán được viết lại thành xy + yz + zx = 3 

Ta có: \(M=\Sigma_{cyc}\frac{a+1}{a^2+2a+2}=\Sigma_{cyc}\frac{a+1}{\left(a+1\right)^2+1}\)\(=\Sigma_{cyc}\frac{1}{a+1+\frac{1}{a+1}}=\Sigma_{cyc}\frac{1}{\frac{\sqrt{3}}{x}+\frac{x}{\sqrt{3}}}\)

\(=\sqrt{3}\left(\frac{x}{x^2+3}+\frac{y}{y^2+3}+\frac{z}{z^2+3}\right)\)

\(=\sqrt{3}\text{​​}\Sigma_{cyc}\left(\frac{x}{x^2+xy+yz+zx}\right)=\sqrt{3}\Sigma_{cyc}\frac{x}{\left(x+y\right)\left(x+z\right)}\)

\(\le\frac{\sqrt{3}}{4}\Sigma_{cyc}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)=\frac{3\sqrt{3}}{4}\)

Đẳng thức xảy ra khi \(x=y=z=1\)hay \(a=b=c=\sqrt{3}-1\)

đây nha bạn

tích mình đi

làm ơn

rùi mình

tích lại

thanks

k mk đi 

Áp dụng BĐT bunhiacopxki ta có :\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(\sqrt{a}.\frac{1}{\sqrt{a}}+\sqrt{b}.\frac{1}{\sqrt{b}}+\sqrt{c}.\frac{1}{\sqrt{c}}\right)^2\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(1+1+1\right)^2=9\)

.Dấu "=" xảy ra khi   :\(\frac{a}{\frac{1}{a}}=\frac{b}{\frac{1}{b}}=\frac{c}{\frac{1}{c}}\Leftrightarrow a^2=b^2=c^2\Leftrightarrow a=b=c\)

Mà \(a+b+c\le\frac{3}{2}\)\(\Rightarrow M=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9:\frac{3}{2}=9.\frac{2}{3}=6\)

Vậy Min M = 6 <=> a = b = c