Chứng minh

Question

$\sqrt{a^2+\frac1{b^2}}+\sqrt{b^2+\frac1{c^2}}+\sqrt{c^2+\frac1{a^2}}\geq \frac{\sqrt{97}}{2}$

với $a,b,c$ là số thực dương thoả mãn $a+b+c\leq2$

score 1 · Answer 1

ta có bdt$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9\rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{2}$ ( vì a+b+c$\leq $2)

áp dụng bdt minkowsky ta có :$A\geq \sqrt{(a+b+c)^{2}+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}=\sqrt{(a+b+c)^{2}+\frac{16}{81}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}+\frac{65}{81}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}\geq \sqrt{2\sqrt{(a+b+c)^{2}.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}.\frac{16}{81}}+\frac{65}{81}.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}\geq \sqrt{2.\sqrt{9^{2}.\frac{16}{81}}+\frac{65}{81}.(\frac{9}{2})^{2}}=\frac{\sqrt{97}}{2}$

dấu bằng xảy ra khi a=b=c=$\frac{2}{3}$

score 3 · Answer 2

ta có bdt$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9\rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{2}$ ( vì a+b+c$\leq $2)

áp dụng bdt minkowsky ta có :$A\geq \sqrt{(a+b+c)^{2}+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}=\sqrt{(a+b+c)^{2}+\frac{16}{81}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}+\frac{65}{81}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}\geq \sqrt{2\sqrt{(a+b+c)^{2}.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}.\frac{16}{81}}+\frac{65}{81}.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}\geq \sqrt{2.\sqrt{9^{2}.\frac{16}{81}}+\frac{65}{81}.(\frac{9}{2})^{2}}=\frac{\sqrt{97}}{2}$

dấu bằng xảy ra khi a=b=c=$\frac{2}{3}$