bài này hên xui

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

\(\ge-a-1-b-1+a+b+\frac{9}{a+b-a+1-b+1}\)

\(=-2+\frac{9}{2}=\frac{5}{2}\)

\(BDT\Leftrightarrow\frac{a^3}{\left(1-a\right)^2}+\frac{b^3}{\left(1-b\right)^2}+\frac{c^3}{\left(1-c\right)^2}\ge\frac{1}{4}\)

Ta có BĐT phụ: \(\frac{a^3}{\left(1-a\right)^2}\ge a-\frac{1}{4}\)

\(\Leftrightarrow\frac{\left(3a-1\right)^2}{4\left(a-1\right)^2}\ge0\forall0< a\le\frac{1}{3}\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\frac{b^3}{\left(1-b\right)^2}\ge b-\frac{1}{4};\frac{c^3}{\left(1-c\right)^2}\ge c-\frac{1}{4}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\left(a+b+c\right)-\frac{1}{4}\cdot3=1-\frac{3}{4}=\frac{1}{4}=VP\)

Xảy ra khi \(a=b=c=\frac{1}{3}\)

Áp dụng BĐT cô si ta có:

\(\frac{a^3}{\left(b+c\right)^2}+\frac{1a}{4}\ge\frac{a^2}{b+c}\)\(,\frac{b^3}{\left(c+a\right)^2}+\frac{1b}{4}\ge\frac{b^2}{a+c},\frac{c^3}{\left(a+b\right)^2}+\frac{1c}{4}\ge\frac{c^2}{a+b}\)

Cộng lại ta có

\(VT\ge\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}-\frac{1}{4}\left(a+b+c\right)\)

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

Dấu =tự tìm Ok

\(A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\left(\frac{x-1}{\sqrt{2}}\right)^2\)

\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\left(x-1\right)^2}{2}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{2}\)

\(=\frac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{2}\)

\(=\frac{x}{\sqrt{x}+1}\cdot\frac{x-1}{2}=\frac{x^2-x}{2\sqrt{x}+2}\)

vào câu hỏi tương tự sẽ có chi tiết hơn nha

C1: \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)

Đặt \(t=\frac{x}{y}+\frac{y}{x}\ge2\Rightarrow t^2=\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\):

\(t^2+2\ge3t\Leftrightarrow\left(t-2\right)\left(t-1\right)\ge0\forall t\ge2\) *đúng*

C2: \(BDT\Leftrightarrow\frac{\left(x-y\right)^2\left(x^2-xy+y^2\right)}{x^2y^2}\ge0\)*đúng*