202601192

$\mathrm{已}\mathrm{知}a>0,b>0,a+b=4，\mathrm{下}\mathrm{列}\mathrm{不}\mathrm{等}\mathrm{式}\mathrm{恒}\mathrm{成}\mathrm{立}\mathrm{的}\mathrm{是}()$

• $A.a^2+b^2>6$
• $B.\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\ge \sqrt{2}$
• $C.a^3+b^3\ge 16$
• $D.\sqrt{a}+\sqrt{b}\ge 2\sqrt{2}$

$\mathrm{对}\mathrm{于}{a}^2+b^2=(a+b{)}^2-2ab=16-2ab\ge 16-2\cdot {\displaystyle \frac{(a+b{)}^2}{4}}\ge 8>6\mathrm{成}\mathrm{立}$
$(a^2+b^2)(1^2+1^2)\ge (a+b{)}^2\Rightarrow a^2+b^2\ge 8$
${\displaystyle \frac{a^2}{1}}+{\displaystyle \frac{b^2}{1}}\ge \frac{(a+b{)}^2}{1+1}=8$
柯西不等式

$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}{)}^2=\frac{1}{a}+\frac{1}{b}+\frac{2}{\sqrt{ab}}=\frac{a+b}{ab}+\frac{2}{\sqrt{ab}}=\frac{4}{ab}+\frac{2}{\sqrt{ab}}$
$\sqrt{ab}\le \frac{a+b}{2}=2,ab\le 4$
$\frac{1}{ab}\ge \frac{1}{4},\frac{1}{\sqrt{ab}}\ge \frac{1}{2}$
$\therefore \frac{4}{ab}+\frac{2}{\sqrt{ab}}\ge 2\ge \sqrt{2}\mathrm{为}\mathrm{真}\mathrm{命}\mathrm{题}$

$b=4-a$
$a^3+b^3=a^3+(4-a{)}^3=12a^2-48a+64=12(a-2{)}^2+16\ge 16$

$(\sqrt{a}+\sqrt{b}{)}^2=a+b+2\sqrt{ab}=4+2\sqrt{ab}\le 4+a+b=8$
$\therefore \sqrt{a}+\sqrt{b}\le 2\sqrt{2}$