ĐKXĐ: $m\neq-1$
$P=\frac{2m^{2}+1}{(m+1)^{2}}$
$=\frac{3(2m^{2}+1)}{3(m+1)^{2}}$
$=\frac{6m^{2}+3)}{3(m+1)^{2}}$
$=\frac{2m^{2}+4m+2+4m^{2}-4m+1}{3(m+1)^{2}}$
$=\frac{2m^{2}+4m+2}{3(m+1)^{2}} +\frac{4m^{2}-4m+1}{3(m+1)^{2}}$
$=\frac{2(m+1)^{2}}{3(m+1)^{2}}+\frac{(2m-1)^{2}}{3(m+1)^{2}}$
$=\frac{2}{3}+\frac{(2m-1)^{2}}{3(m+1)^{2}}$
⇒$P\geq\frac{2}{3}$ với ∀$m\neq-1$ (vì $(2m-1)^{2}\geq0$)
Dấu "=" xảy ra⇔$2m-1=0$
⇔$m=\frac{1}{2}$ (t/m)
Vậy Min$P=\frac{2}{3}$ khi $m=\frac{1}{2}$
