BĐT Bunhiacopski cho 3 số$(a+b+c)(\frac{x^{2}}{a}+\frac{y^{2}}{b}+\frac{z^{2}}{c})\geq(x+y+z)^{2}$(*)
có$P=\frac{a^{4}}{ab+2ac}+\frac{b^{4}}{bc+2ab}+\frac{c^{4}}{ac+2bc}$
áp dụg BĐT (*) ta có$ (ab+2ac+bc+2ab+bc+2bc)P\geq(a^{2}+b^{2}+c^{2})^{2}$
hay $P\geq\frac{(a^{2}+b^{2}+c^{2})^{2}}{(3ab+3bc+3ac)}\geq\frac{1}{9}$
dấu "=" xảy ra <=>$a=b=c=\frac{1}{3}$