Câu hỏi của Đẹp Trai Không Bao Giờ Sai

Question

Cho $a;b\ge0$ thỏa mãn $a^2+b^2=a+b$Tìm GTLN của $S=\frac{a}{a+1}+\frac{b}{b+1}$

Phan Công Bằng · Accepted Answer

Có $\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0$

$\Rightarrow2a^2+2b^2\ge a^2+2ab+b^2\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2$

$\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge\frac{\left(a+b\right)^2}{2}$

$\Rightarrow a+b\ge\frac{\left(a+b\right)^2}{2}\Rightarrow2\ge a+b$

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

AD BĐT: $\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x,y\in Z^+\right)$

$\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+b+2}\ge\frac{4}{4}=1$ ( vì $2\ge a+b$ )

$\Rightarrow S=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\le1$

Dấu "=" xảy ra $\Leftrightarrow a=b=1$

Vậy $S_{max}=1\Leftrightarrow a=b=1$Câu trả lời: Có $\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0$