Đáp án:
$(a+b)(b+c)(c+a)\le\dfrac{1}{8}$
Giải thích các bước giải:
$\text{Đặt a+b+1=x, b+c+1=y, c+a+1=z}$
$\text{ Khi đó đề bài trở thành cho x,y,z dương thỏa mãn:}$
$\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=2(1)$
$\text{Tìm max P=xyz}$
$\text{Từ (1) ta có:}$
$(x+1)(y+1)+(y+1)(z+1)+(z+1)(x+1)=2(x+1)(y+1)(z+1)$
$\rightarrow (xy+yz+zx)+2(x+y+z)+3=2xyz+2+2(xy+yz+zx)+2(x+y+z)$
$\rightarrow 1=2xyz+(xy+yz+zx)\ge 2xyz+3\sqrt[3]{xy.yz.zx}$
$\rightarrow 2xyz+3(\sqrt[3]{xyz})^2-1 \le 0$
$\rightarrow (2\sqrt[3]{xyz}-1)(\sqrt[3]{xyz}+1)^2\le 0$
$\rightarrow 2\sqrt[3]{xyz}-1\le 0$
$\rightarrow \sqrt[3]{xyz}\le \dfrac{1}{2}$
$\rightarrow xyz\le\dfrac{1}{8}$
$\text{Hay }(a+b)(b+c)(c+a)\le\dfrac{1}{8}$
