Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Đặt a+b=x;b+c=y;c+a=z

\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)

Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)

\(S=\frac{a}{1+b}+\frac{b}{1+a}+\frac{1}{a+b}=\frac{a^2}{a+ab}+\frac{b^2}{b+ab}+\frac{1}{a+b}\)

\(S\ge\frac{\left(a+b\right)^2}{a+b+2ab}+\frac{1}{a+b}\ge\frac{\left(a+b\right)^2}{a+b+\frac{\left(a+b\right)^2}{2}}+\frac{1}{a+b}\)

\(S\ge\frac{2\left(a+b\right)}{a+b+2}+\frac{1}{a+b}=2-\frac{4}{a+b+2}+\frac{1}{a+b}\)

Đặt \(a+b=t\Rightarrow0< t\le1\)

\(S\ge\frac{5}{3}+\frac{t+3}{3t}-\frac{4}{t+2}=\frac{5}{3}+\frac{t^2-7t+6}{3t\left(t+2\right)}=\frac{5}{3}+\frac{\left(6-t\right)\left(1-t\right)}{3t\left(t+2\right)}\ge\frac{5}{3}\)

\(S_{min}=\frac{5}{3}\) khi \(t=1\Leftrightarrow x=y=\frac{1}{2}\)

từ giả thiết, ta có \(\frac{a^2}{b}+\frac{b^2}{a}\le1\)

Mà \(\frac{a^2}{b}+\frac{b^2}{a}\ge\frac{\left(a+b\right)^2}{a+b}=a+b\Rightarrow a+b\le1\)

Mà từ BĐT cô-si, ta luôn có \(\left(a+b\right)^3\ge4ab\left(a+b\right)\ge4\left(a^3+b^3\right)\left(a+b\right)\Rightarrow\frac{\left(a+b\right)^3}{4}\ge\left(a^3+b^3\right)\left(a+b\right)\)

Mà áp dụng BĐT Bu-nhi-a , ta có \(\left(a^3+b^3\right)\left(a+b\right)\ge\left(a^2+b^2\right)^2\)

=>\(\frac{\left(a+b\right)^3}{4}\ge\left(a^2+b^2\right)^2\Rightarrow\frac{1}{4}\ge\left(a^2+b^2\right)^2\Rightarrow a^2+b^2\le\frac{1}{2}\)

Mà \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+a^2+b^2}=\frac{4}{2+\frac{1}{2}}=\frac{8}{5}\)

Dấu = xảy ra ,=> a=b=1/2

^_^

\(a^3+b^3\le ab\Leftrightarrow ab\left(a+b\right)\le ab\Leftrightarrow a+b\le1.\).Ta có: \(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}.\)

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+a^2+b^2}=\frac{4}{2+\left(a+b\right)^2-2ab}\ge\frac{4}{2+1-\frac{1}{2}}\ge\frac{8}{5}.\)

Dấu bằng xảy ra khi a=b=1/2.

Trần Hữu Ngọc Minh, cho t hỏi 

\(ab\le\frac{1}{4}\Rightarrow-2ab\ge-\frac{1}{2}\)

Tức là k đánh giá được mẫu !!!?

Mình học lớp 8 nên vẫn chưa biết "Min" là gì vậy bạn?

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

\(=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+4\)

Dễ có:\(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)

\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\ge\frac{2}{\frac{\left(a+b\right)^2}{4}}=\frac{8}{\left(a+b\right)^2}=8\)

Khi đó:\(S\ge\frac{1}{2}+8+4=\frac{25}{2}\)

Vậy ta có đpcm

\(S\ge\frac{\left(a+b+\frac{4}{a+b}\right)^2}{2}=\frac{25}{2}\)

Đẳng thức xảy ra khi a = b = \(\frac{1}{2}\)