Đáp án: $a=2, b=4, c=6$
Giải thích các bước giải:
Ta có:
$\dfrac1a+\dfrac4b+\dfrac9c=3$
$\to 3(a+b+c)= (\dfrac1a+\dfrac4b+\dfrac9c)(a+b+c)$
$\to 3(a+b+c)\ge (\sqrt{ \dfrac1a\cdot a}+\sqrt{\dfrac4b\cdot b}+\sqrt{\dfrac9c\cdot c})^2$ (BĐT bunhiacopxki)
$\to 3(a+b+c)\ge (1+2+3)^2$
$\to 3(a+b+c)\ge 6^2$
$\to a+b+c\ge 12$
Mà $a+b+c\le 12\to a+b+c=12$
Dấu = xảy ra khi $\dfrac1a:a=\dfrac4b:b=\dfrac9c:c$
$\to \dfrac1{a^2}=\dfrac4{b^2}=\dfrac9{c^2}$
$\to \dfrac1a=\dfrac2b=\dfrac3c=\dfrac{1+2+3}{a+b+c}=\dfrac6{12}=\dfrac12$
$\to a=2, b=4, c=6$
