Question

Cho các số a, b, c thỏa mãn

\(a+b+c=abc\)

Tìm Max S = \(\frac{a}{\sqrt{bc.\left(1+a^2\right)}}+\frac{b}{\sqrt{ac.\left(1+b^2\right)}}+\frac{c}{\sqrt{ab.\left(1+c^2\right)}}\)

Answer 1

a b c la : nhau vay a 2 b 5 c 9

Answer 2

dap an laf a 4  b 6c 14

Answer 3

Nêu rõ cách làm đi bạn ơi

Answer 4

cach lam a 

Answer 5

\(S=\frac{a}{\sqrt{bc\left(1+a^2\right)}}+\frac{b}{\sqrt{ac\left(1+b^2\right)}}+\frac{c}{\sqrt{ab\left(1+c^2\right)}}\)

\(=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}+\frac{b}{\sqrt{ac+b\left(a+b+c\right)}}+\frac{c}{\sqrt{ab+c\left(a+b+c\right)}}\)

\(=\frac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)

Xét: \(\left(\frac{a}{a+b}+\frac{a}{a+c}\right)^2=\frac{a^2}{\left(a+b\right)^2}+\frac{a^2}{\left(b+c\right)^2}+2\frac{a^2}{\left(a+b\right)\left(b+c\right)}\ge\frac{4a^2}{\left(a+b\right)\left(b+c\right)}\)(a,b cùng dấu)

\(\Rightarrow\frac{\frac{a}{a+b}+\frac{a}{a+c}}{2}\ge\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

Áp dụng tương tự, ta có:\(S\le\frac{\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{a+c}{a+c}}{2}=\frac{3}{2}\)

Dấu = xảy ra khi \(a=b=c=\pm\sqrt{3}\)