Trích:
Nguyên văn bởi bac_hai_60 Bài 2 : Cho ba số a,b,c thỏa mãn $\[a,b,c>0,a+b+c=3\] $. Chứng minh rằng : $\[\sqrt{2{{a}^{2}}+\frac{2}{a+1}+{{b}^{4}}}+\sqrt{2{ {b}^{2}}+\frac{2}{b+1}+{{c}^{4}}}+\sqrt{2{{c}^{2}} +\frac{2}{c+1}+{{a}^{4}}}\ge 6\] $ |
Áp dụng bất đẳng thức Minkowski, ta được
$\[\sqrt{2{{a}^{2}}+\frac{2}{a+1}+{{b}^{4}}}+\sqrt{2{ {b}^{2}}+\frac{2}{b+1}+{{c}^{4}}}+\sqrt{2{{c}^{2}} +\frac{2}{c+1}+{{a}^{4}}}\ge \sqrt{2(a+b+c)^2+2(\frac{1}{\sqrt{a+1}}+\frac{1}{ \sqrt{b+1}} +\frac{1}{\sqrt{c+1}})^2+(a+b+c)^2}=\sqrt{27+2( \frac{1}{ \sqrt{a+1}}+\frac{1}{ \sqrt{b+1}} +\frac{1}{\sqrt{c+1}})^2} \ge \sqrt{27+2(\frac{3}{ \sqrt[6]{(a+1)(b+1)(c+1)}})^2 $$= \sqrt{27+ \frac{18}{\sqrt[3]{(a+1)(b+1)(c+1)}}} \ge \sqrt{27+\frac{54}{a+1+b+1+c+1}}=\sqrt{36}=6 $
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]