Question

Answer 1

Lời giải:

Áp dụng BĐT AM-GM: $1\geq a+b\geq 2\sqrt{ab}$

$\Rightarrow ab\leq \frac{1}{4}$

Áp dụng BĐT Cauchy-Schwarz:

$D=\frac{1}{1+a^2+b^2}+\frac{1}{6ab}+\frac{1}{6ab}+\frac{1}{6ab}$

$\geq \frac{(1+1+1+1)^2}{1+a^2+b^2+6ab+6ab+6ab}$

$=\frac{16}{1+(a+b)^2+16ab}$
$\geq \frac{16}{1+1^2+16.\frac{1}{4}}=\frac{8}{3}$
Vậy $D_{\min}=\frac{8}{3}$ khi $a=b=\frac{1}{2}$