Theo Cauchy:
$a+(b+c)\geq 2\sqrt{a(b+c)}\Leftrightarrow \sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}$
Tương tự:
$\sqrt{\frac{b}{a+c}}\geq \frac{2b}{a+b+c}$
$\rightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\geq \frac{2(a+b)}{a+b+c}$
Do đó:
$P\geq \frac{2(a+b)}{a+b+c}+\frac{c}{2(a+b)}=[\frac{2(a+b)}{a+b+c}+\frac{a+b+c}{2(a+b)}]-\frac{1}{2}$
$\rightarrow P\geq 2-\frac{1}{2}=\frac{3}{2}$
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