Question

Cho a,b,c > 0 , \(a^2+b^2+c^2=3\). Chứng minh rằng : \(\sqrt{\frac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\frac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\frac{9}{\left(a+c\right)^2}+b^2}\)≥\(\frac{3\sqrt{13}}{2}\)

Answer 1

https://hoc24.vn/hoi-dap/question/620649.html

Answer 2

Kết hợp Mincôpxki và C-S:

\(VT\ge\sqrt{\left(\frac{3}{a+b}+\frac{3}{b+c}+\frac{3}{a+c}\right)^2+\left(a+b+c\right)^2}\)

\(VT\ge\sqrt{\left(\frac{27}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}=\sqrt{\frac{405}{4\left(a+b+c\right)^2}+\frac{81}{\left(a+b+c\right)^2}+\left(a+b+c\right)^2}\)

\(VT\ge\sqrt{\frac{405}{12\left(a^2+b^2+c^2\right)}+2\sqrt{\frac{81\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}}=\sqrt{\frac{405}{12.3}+18}=\frac{3\sqrt{13}}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)