cho $\left\{\begin{matrix} a,b,c >0 & & \\ a+b=c\leq 1& & \end{matrix}\right.$
tìm GTNN
$b, \frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{a^{2}+c^{2}}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}$
$b, \frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{a^{2}+c^{2}}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}$
$=(\frac{1}{a^2+b^2}+\frac{1}{2ab})+(\frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}+\frac{(\frac{2}{5})^2}{2ab})+(\frac{1}{ac}+\frac{1}{bc})+\frac{21}{50ab}$
$\geq \frac{4}{a^2+b^2+2ab}+\frac{(1+1+\frac{2}{5})^2}{a^2+b^2+2c^2+2ab}+\frac{4}{ac+bc}+\frac{42}{25(a+b)^2}$
$=\frac{4}{(a+b)^2}+\frac{144}{25[(a+b)^2+2c^2]}+\frac{4}{c(a+b)}+\frac{42}{25(a+b)^2}$
$=\frac{4}{c^2}+\frac{48}{25c^2}+\frac{4}{c^2}+\frac{42}{25c^2}=\frac{38}{5c^2} \geq \frac{38}{5}$
Dấu $=$ xảy ra khi và chỉ khi $a=b=\frac{1}{2}, c=1$.