Lời giải:
Theo công thức Herong thì:
\(S=\sqrt{p(p-a)(p-b)(p-c)}=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\)
Do vậy ta cần CM: \(\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\leq \frac{\sqrt{3}}{4}.\sqrt[3]{a^2b^2c^2}\)
\(\Leftrightarrow (a+b+c)^3(a+b-c)^3(b+c-a)^3(c+a-b)^3\leq 27(abc)^4\)
Đặt \(\left\{\begin{matrix} a+b-c=x\\ b+c-a=y\\ c+a-b=z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a=\frac{x+z}{2}\\ b=\frac{x+y}{2}\\ c=\frac{y+z}{2}\end{matrix}\right.\)
Điều cần CM trở thành: \(\frac{4096}{27}(x+y+z)^3(xyz)^3\leq [(x+y)(y+z)(x+z)]^4\)
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Thật vậy:
Ta có bổ đề quen thuộc: \((x+y)(y+z)(x+z)\geq \frac{8}{9}(xy+yz+xz)(x+y+z)\)
\(\Rightarrow [(x+y)(y+z)(x+z)]^4\geq \frac{4096}{9^4}(xy+yz+xz)^4(x+y+z)^4\)
Mà theo BĐT AM-GM:
\( \frac{4096}{9^4}(xy+yz+xz)^4(x+y+z)^4=\frac{4096}{27}(x+y+z)^3.\frac{(xy+yz+xz)^4(x+y+z)}{243}\)
\(\geq \frac{4096}{27}(x+y+z)^3.\frac{(3\sqrt[3]{x^2y^2z^2}]^4.3\sqrt[3]{xyz}}{243}=\frac{4066}{27}(x+y+z)^3(xyz)^3\)
Do đó: \([(x+y)(y+z)(x+z)]^4\geq \frac{4066}{27}(x+y+z)^3(xyz)^3\) (đpcm)
Vậy............
