Khai triển Logarit:$In S=\Sigma [b*In (a+\sqrt{a^{2}+1})]$
Xét hàm số $f(x)=In(x+\sqrt{x^{2}+1}),x>0$,ta có:
$f'(x)=\frac{1}{\sqrt{x^{2}+1}}; f''(x)=-\frac{x}{(x^{2}+1)\sqrt{x^{2}+1}}<0$
$\Rightarrow f(x)\leq f'(\frac{3}{4})(x-\frac{3}{4})+f(\frac{3}{4})=\frac{4}{5}x+In2-\frac{3}{5}$(Use BĐT tiếp tuyến)
$\Rightarrow In S \leq \frac{4}{5}(ab+bc+ca)+(In 2-\frac{9}{5})(a+b+c)\le \frac{27}{20}+In 2*\frac{9}{4}-\frac{27}{20}=ln2*\frac{9}{4}$
$\Rightarrow S\le e^{\frac{9}{4}ln(2)}$
Dấu''='' xra$\Leftrightarrow a=b=c=\frac{3}{4}$