Câu hỏi của dam thu a

Question

cho a,b >0 thỏa mãn $\left(a+1\right)\left(b+1\right)=\frac{9}{4}$ . tìm GTNN của $A=\sqrt{a^4+1}+\sqrt{b^4+1}$

Nguyễn Việt Lâm · Accepted Answer

$ab+a+b=\frac{5}{4}\Rightarrow\frac{a^2+b^2}{2}+\sqrt{2\left(a^2+b^2\right)}\ge\frac{5}{4}$

$\Rightarrow a^2+b^2\ge\frac{1}{2}$

$A=\sqrt{a^4+1}+\sqrt{b^4+1}\ge\sqrt{\left(a^2+b^2\right)^2+4}\ge\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}$

Dấu "=" xảy ra khi $a=b=\frac{1}{2}$Câu trả lời: $ab+a+b=\frac{5}{4}\Rightarrow\frac{a^2+b^2}{2}+\sqrt{2\left(a^2+b^2\right)}\ge\frac{5}{4}$

$\Rightarrow a^2+b^2\ge\frac{1}{2}$

$A=\sqrt{a^4+1}+\sqrt{b^4+1}\ge\sqrt{\left(a^2+b^2\right)^2+4}\ge\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}$

Dấu "=" xảy ra khi $a=b=\frac{1}{2}$