Question

Cho \(-1\le a\le1\). Tìm GTLN của b sao cho BĐT đúng \(\sqrt{1-a^4}+\left(b+1\right)\left(\sqrt{1+a^2}+\sqrt{1-a^2}\right)+b-4\le0\)

Answer 1

Đặt \(\sqrt{1+a^2}+\sqrt{1-a^2}=x\Rightarrow\sqrt{2}\le x\le2\)

\(x^2=2+2\sqrt{1-a^4}\Rightarrow\sqrt{1-a^4}=\dfrac{x^2-2}{2}\)

\(\Rightarrow\dfrac{x^2-2}{2}+\left(b+1\right)x+b-4\le0\)

\(\Rightarrow x^2+2\left(b+1\right)x+2b-10\le0\)

\(\Rightarrow x^2+2x-10\le-2b\left(x+1\right)\)

\(\Rightarrow-2b\ge\dfrac{x^2+2x-10}{x+1}\)

\(\Rightarrow-2b\ge\max\limits_{\left[\sqrt{2};2\right]}f\left(x\right)\) với \(f\left(x\right)=\dfrac{x^2+2x-10}{x+1}\)

Xét trên \(\left[\sqrt{2};2\right]\) ta có:

\(f\left(x\right)=\dfrac{3x^2+6x-30}{3\left(x+1\right)}=\dfrac{3x^2+8x-28-2\left(x+1\right)}{3\left(x+1\right)}=\dfrac{\left(3x+14\right)\left(x-2\right)}{3\left(x+1\right)}-\dfrac{2}{3}\le-\dfrac{2}{3}\)

\(\Rightarrow-2b\ge-\dfrac{2}{3}\Rightarrow b\le\dfrac{1}{3}\)

Vậy \(b_{max}=\dfrac{1}{3}\)