cách 1.có $\mathbb {P}= (a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})-3\geq(a+b+c)\frac{9}{2(a+b+c)}-3=\frac{3}{2}$
suy ra DPCM
cách 2 có
$P=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+ba}+\frac{c^{2}}{ca+cb}\geq\frac{(a+b+c)^{2}}{2(ab+bc+ca)}$ (Bu nhi a)
$\geq\frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac{3}{2}$
suy ra DPCM