Đáp án: $N\le \dfrac12+\dfrac{1}{2\sqrt2}+\dfrac1{2\sqrt3}$
Giải thích các bước giải:
Ta có :
$\sqrt{c-4}=\dfrac12\sqrt{4(c-4)}\le\dfrac12.\dfrac{4+c-4}{2}=\dfrac{c}{2}$
$\to ab\sqrt{c-4}\le \dfrac{abc}{2}$
Dấu = xảy ra khi $4=c-4\to c=8$
Lại có :
$\sqrt{a-2}=\dfrac{1}{\sqrt{2}}.\sqrt{2(a-2)}\le \dfrac{1}{\sqrt{2}}\dfrac{2+a-2}{2}=\dfrac{a}{2\sqrt{2}}$
$\to bc\sqrt{a-2}\le \dfrac{abc}{2\sqrt2}$
Dấu = xảy ra khi $2=a-2\to a=4$
$\sqrt{b-3}=\dfrac{1}{\sqrt{3}}.\sqrt{3(b-3)}\le \dfrac{1}{\sqrt{3}}.\dfrac{3+b-3}{2}=\dfrac{b}{2\sqrt{3}}$
$\to ca\sqrt{b-3}\le \dfrac{abc}{2\sqrt{3}}$
Dấu = xảy ra khi $3=b-3\to b=6$
$\to N\le \dfrac12+\dfrac{1}{2\sqrt2}+\dfrac1{2\sqrt3}$
Dấu = xảy ra khi $a=4,b=6,c=8$
