Cho ba số dương a,b,c>0.
2)Cho $a+b+c+\sqrt{2abc}\geq 10$ . Chứng minh rằng:
$\sqrt{\frac{8}{a^{2}}+\frac{9b^{2}}{2}+\frac{c^{2}a^{2}}{4}}+\sqrt{\frac{8}{b^{2}}+\frac{9c^{2}}{2}+\frac{a^{2}b^{2}}{4}}+\sqrt{\frac{8}{c^{2}}+\frac{9a^{2}}{2}+\frac{b^{2}c^{2}}{4}}\geq 6\sqrt{6}$.
3)Cho a+b+c=3. Chứng minh:
$\sqrt{2a^{2}+\frac{2}{a+1}+b^{4}}+\sqrt{2b^{2}+\frac{2}{b+1}+c^{4}}+\sqrt{2c^{2}+\frac{2}{c+1}+a^{4}}\geq 6$.
Mình xin giải 2 bài cuối:
Bài 2) Áp dụng BĐT Cauchy-Schwarz ta có:
$\sum \sqrt{(\frac{8}{a^2}+\frac{9}{b^2}+\frac{c^2a^2}{4})(2+18+4))} \geq \sum (\frac{4}{a}+9b+ca)$
$=> \sum \sqrt{(\frac{8}{a^2}+\frac{9}{b^2}+\frac{c^2a^2}{4})}\geq \frac{1}{\sqrt{24}}(4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+9(a+b+c)+ab+bc+ca)\geq \frac{1}{\sqrt{24}}(\frac{36}{a+b+c}+9(a+b+c)+ab+bc+ca)= \frac{1}{\sqrt{24}}(\frac{36}{a+b+c}+a+b+c+6(a+b+c)+2(a+b+c)+ab+bc+ca)\geq \sqrt{24}(12+6(a+b+c)+6\sqrt{2abc}))(AM-GM) \geq \frac{1}{\sqrt{24}}(12+6.10)(gt)= 6\sqrt{6}$
Đẳng thức xảy ra khi $a=b=c=2$
Bài 3) Áp dụng BĐT Cauchy-Schwarz ta có:
$\sum \sqrt{(2a^2+\frac{2}{a+1}+b^4)(2+1+1)}\geq \sum (2a+\sqrt{\frac{2}{a+1}}+b^2)$
$=> VT\geq \frac{1}{2}\sum (2a+\sqrt{\frac{2}{a+1}}+b^2)= \frac{1}{2}(\frac{1}{2}\sum (\sqrt{\frac{2}{a+1}}+\sqrt{\frac{2}{a+1}}+\frac{a+1}{2})+(\frac{7}{4}(a+b+c)-\frac{3}{4})+(a^2+1)+(b^2+1)+(c^2+1)-3))\geq \frac{1}{2}(\frac{9}{2}+\frac{21}{4}-\frac{3}{4}+6-3)=6$
Đẳng thức xảy ra khi $a=b=c=1$
Bài viết đã được chỉnh sửa nội dung bởi nghiakvnvsdt: 23-10-2013 - 21:09
