Question

Cho a,b>0 thỏa mãn a + b + 3ab = 1. Tìm GTLN P = \(\sqrt{1-a^2}+\sqrt{1-b^2}+\dfrac{3ab}{a+b}\)

Answer 1

Lời giải:

$1=a+b+3ab\leq (a+b)+3.\frac{(a+b)^2}{4}$

$\Rightarrow a+b\geq \frac{2}{3}$

$\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}=\frac{2}{9}$

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

\(\leq \sqrt{(1-a^2+1-b^2)(1+1)}+\frac{1}{\frac{2}{3}}-1=\sqrt{2(2-a^2-b^2)}+\frac{1}{2}\)

Mà \(2-a^2-b^2\leq 2-\frac{2}{9}=\frac{16}{9}\)

Do đó:

\(P\leq \sqrt{\frac{32}{9}}+\frac{1}{2}=\frac{3+8\sqrt{2}}{6}\) và đây chính là giá trị max.