Lời giải:
Áp dụng BĐT AM-GM:
\(P=\frac{\sqrt{ab}}{(a+c)+(b+c)}+\frac{\sqrt{bc}}{(b+a)+(c+a)}+\frac{\sqrt{ca}}{(c+b)+(a+b)}\)
\(\leq \underbrace{\frac{\sqrt{ab}}{2\sqrt{(a+c)(b+c)}}+\frac{\sqrt{bc}}{2\sqrt{(b+a)(c+a)}}+\frac{\sqrt{ca}}{2\sqrt{(c+b)(a+b)}}}_{M}(*)\)
Xét:
\(M=\frac{1}{2}\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}(1)\)
Theo BĐT Bunhiacopxky và AM-GM:
\((\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)})^2\leq (ab+bc+ac)(a+b+b+c+c+a)\)
\(=2(ab+bc+ac)(a+b+c)=2[(a+b)(b+c)(c+a)+abc]\)
\(\leq 2[(a+b)(b+c)(c+a)+\frac{(a+b)(b+c)(c+a)}{8}]=\frac{9}{4}(a+b)(b+c)(c+a)\)
\(\Rightarrow \sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}\leq \frac{3}{2}\sqrt{(a+b)(b+c)(c+a)}(2)\)
Từ \((1);(2)\Rightarrow M\leq \frac{1}{2}.\frac{3}{2}=\frac{3}{4}(**)\)
Từ \((*); (**)\Rightarrow P\leq M\leq \frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow a=b=c\)
2\(\dfrac{\sqrt{ab}}{a+b+2c},< =\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}< =\dfrac{a}{a+c}+\dfrac{b}{b+c}\)
tương tự
=> 2P≤3
