Giải thích các bước giải:
Ta có:
$\dfrac{a}{1+a}+\dfrac{2b}{1+b}+\dfrac{3c}{1+c}\le 1$
$\to a(1+b)(1+c)+2b(1+a)(1+c)+3c(1+a)(1+b)\le (1+a)(1+b)(1+c)$
$\to 3ab+4ac+6abc+a+2b+5bc+3c\le 1+c+b+bc+a+ac+ab+abc$
$\to 2ab+3ac+5abc+b+4bc+2c\le 1$
$\to 5(2ab+3ac+5abc+b+4bc+2c)\le 5$
$\to 5\ge 10ab+15ac+25abc+5b+20bc+10c$
$\to 5\ge (5ab+5ab)+(5ac+5ac+5ac)+25abc+(b+b+b+b+b)+(5bc+5bc+5bc+5bc)+(c+c+c+c+c+c+c+c+c+c)$
$\to 5\ge 5ab+5ab+5ac+5ac+5ac+25abc+b+b+b+b+b+5bc+5bc+5bc+5bc+c+c+c+c+c+c+c+c+c+c$
$\to 5\ge 25\sqrt[25]{5ab\cdot 5ab\cdot 5ac\cdot 5ac\cdot 5ac\cdot 25abc\cdot b\cdot b\cdot b\cdot b\cdot b\cdot 5bc\cdot 5bc\cdot 5bc\cdot 5bc\cdot c\cdot c\cdot c\cdot c\cdot c\cdot c\cdot c\cdot c\cdot c\cdot c}$
$\to 5\ge 25\sqrt[25]{5^{11}\cdot a^6\cdot b^{12}\cdot c^{18}}$
$\to ab^2c^3\le \dfrac1{5^6}$
Dấu = xảy ra khi $a=b=c=\dfrac15$
