Theo bđt cô si ta có: $√xy≤$$\dfrac{x+y}{2}$
⇒$xy≤$$\dfrac{(x+y)^2}{4}$
Dấu = xảy ra ⇔$x=y$
Áp dụng bđt này ta có:
$ab≤$$\dfrac{(a+b)^2}{4}$
⇒$\dfrac{ab}{a+b}$$≤$$\dfrac{(a+b)^2}{4}$$:(a+b)=$$\dfrac{a+b}{4}$
$bc≤$$\dfrac{(b+c)^2}{4}$
⇒$\dfrac{bc}{b+c}$$≤$$\dfrac{(b+c)^2}{4}$$:(c+b)=$$\dfrac{c+b}{4}$
$ca≤$$\dfrac{(a+c)^2}{4}$
⇒$\dfrac{ac}{a+c}$$≤$$\dfrac{(a+c)^2}{4}$$:(a+c)=$$\dfrac{a+c}{4}$
⇒$\dfrac{ab}{a+b}$+$\dfrac{bc}{b+c}$+$\dfrac{ac}{a+c}$ $\leq$$\dfrac{a+b}{4}$+$\dfrac{c+b}{4}$+$\dfrac{a+c}{4}$=$\dfrac{2(a+b+c)}{4}$=$\dfrac{a+b+c}{2}$
⇒$đpcm$
Dấu = xảy ra ⇔$a=b=c$
