Cho $a,b,c\geq 0,a\geq c,b\geq c$. Tìm Min :
$P=\frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}+\sqrt{a+b+c}$
Ta có ,do $c\leq a= > a^2+c^2\leq a^2+ac=(a+\frac{c}{2})^2-\frac{c^2}{4}\leq (a+\frac{c}{2})^2$
$c\leq b= > b^2+c^2\leq b^2+bc=(b+\frac{c}{2})^2-\frac{c^2}{4}\leq (b+\frac{c}{2})^2$
Từ đó $= > \frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}\geq \frac{1}{(a+\frac{c}{2})^2}+\frac{1}{(b+\frac{c}{2})^2}\geq \frac{8}{(a+\frac{c}{2}+b+\frac{c}{2})^2}=\frac{8}{(a+b+c)^2}$
(Do áp dụng bất đẳng thức$\frac{1}{m^2}+\frac{1}{n^2}\geq \frac{8}{(m+n)^2}$. Chứng minh bằng Cosi thì $\frac{1}{m^2}+\frac{1}{n^2}\geq \frac{2}{mn}=\frac{8}{4mn}\geq \frac{8}{(m+n)^2}$)
Từ đó $= > P\geq \frac{8}{(a+b+c)^2}+\sqrt{a+b+c}=\frac{8}{(a+b+c)^2}+\frac{\sqrt{a+b+c}}{4}+\frac{\sqrt{a+b+c}}{4}+\frac{\sqrt{a+b+c}}{4}+\frac{\sqrt{a+b+c}}{4}\geq 5\sqrt[5]{\frac{8\sqrt{(a+b+c)^4}}{4^4.(a+b+c)^2}}=5\sqrt[5]{\frac{8(a+b+c)^2}{4^4(a+b+c)^2}}=5\sqrt[5]{\frac{8}{4^4}}=\frac{5}{2}= > P\geq \frac{5}{2}= > P_{min}=\frac{5}{2}< = > a=b,c=0, a+b+c=4< = > a=b=2,c=0$