Chủ đề này có 1 trả lời

[qazplm654]

1. Cho $a,b,c>0; a+b+c\geq \frac{3}{2}$. Tìm Min $A=a^2+b^2+c^2+\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}$.

2. Cho $a,b,c>0; a+b+c \leq \frac{3}{2}$. Tìm Min $B=\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}$

3. Cho a,b,c>0, a=Max{a,b,c}. Tìm Min $C=\frac{a}{b}+2\sqrt{1+\frac{b}{c}}+3\sqrt[3]{1+\frac{c}{a}}$.

4. Cho $T=[\frac{a}{(a-b)(ab-1)}]^2+[\frac{b}{(a-b)(a^2-1)}]^2+[\frac{a^3b}{(a^2-1)(ab-1)}]^2$. Khi T xác định, cm $T^2+3T^{-1} \geq 10$

[Duy Thai2002]

2)Bổ đề: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}$ 

Ta có: $B=\sum \sqrt{a^{2}+\frac{1}{b^{2}}}\geq \sum \sqrt{\frac{1}{17}}(a+\frac{4}{b})$(Cauchy-Schwarz)=$\sqrt{\frac{1}{17}}((\sum a+\frac{1}{4a})+\frac{15}{4a}+\frac{15}{4b}+\frac{15}{4c})\geq \sqrt{\frac{1}{17}}(\sum 2\sqrt{a.\frac{1}{4a}}+\frac{15}{4}.\frac{9}{a+b+c})\geq\sqrt{\frac{1}{17}}.(3+\frac{15}{4}.\frac{9}{\frac{3}{2}})=\frac{3\sqrt{17}}{2}$

=> $B\geq \frac{3\sqrt{17}}{2}.$ĐTXR $<=> x=y=z=\frac{1}{2}$

Vậy Min B= $\frac{3\sqrt{17}}{2} <=> x=y=z=\frac{1}{2}$

Bài viết đã được chỉnh sửa nội dung bởi Duy Thai2002: 16-07-2017 - 23:03