Lời giải:
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 2(a^2+b^2)\geq (a+b)^2\)
\(a+b=a^2+b^2\geq \frac{(a+b)^2}{2}\Rightarrow 2(a+b)\geq (a+b)^2\)
Do đó mà \(a+b\leq 2\)
Có:
\(S=\frac{a}{a+1}+\frac{b}{b+1}=1-\frac{1}{a+1}+1-\frac{1}{b+1}=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a+1}+\frac{1}{b+1}\geq \frac{4}{a+b+2}\geq \frac{4}{2+2}=1\) do \(a+b\leq 2\)
Do đó: \(S\leq 2-1\Leftrightarrow S\leq 1\)
Vậy \(S_{\max}=1\Leftrightarrow a=b=1\)
\(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}=\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}\)
Áp dụng BĐT Cauchy-Schwarz, ta có:
\(\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}\ge\dfrac{\left(a+b\right)^2}{a^2+b^2+a+b}=\dfrac{\left(a+b\right)^2}{2\left(a^2+b^2\right)}\)
Bởi vì:\(a^2+b^2=a+b\Rightarrow a^2+b^2+a+b=a^2+b^2+a^2+b^2=2\left(a^2+b^2\right)\)
Mặt khác, theo Bunyakovsky, ta có:
\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
Do đó: \(\dfrac{\left(a+b\right)^2}{2\left(a^2+b^2\right)}\le\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2}=1\)
Vậy: \(Max_S=1\Leftrightarrow a=b=1\)
