Question

Chứng minh \(\frac{a}{\sqrt{5a^2+\left(b+c\right)^2}}+\frac{b}{\sqrt{5b^2+\left(a+c\right)^2}}+\frac{c}{\sqrt{5c^2+\left(b+a\right)^2}}\le1\)với a,b,c thực dương

Giúp em vs ạ!

Answer 1

a,b,c thực dương

Answer 2

dễ thế mà ko làm dc à ngu vậy má

Answer 3

a,b,c thuộc rỗng

Answer 4

Đặt: a + b + c = t  ta có:

\(5a^2+\left(b+c\right)^2=5a^2+\left(t-a\right)^2=6a^2-2at+t^2=\frac{5}{9}\left(t-3a\right)^2+\left(a+\frac{2}{3}t\right)^2\)

\(\ge\left(a+\frac{2}{3}t\right)^2\)

=> VT \(\le\frac{a}{a+\frac{2}{3}t}+\frac{b}{b+\frac{2}{3}t}+\frac{c}{c+\frac{2}{3}t}\)

\(=\frac{3a}{5a+2b+2c}+\frac{3b}{2a+5b+2c}+\frac{3c}{2a+2b+5c}\)

\(=\left(1-\frac{2a+2b+2c}{5a+2b+2c}\right)+\left(1-\frac{2a+2b+2c}{2a+5b+2c}\right)+\left(1-\frac{2a+2b+2c}{2a+2b+5c}\right)\)

\(=3-2\left(a+b+c\right)\left(\frac{1}{5a+2b+2c}+\frac{1}{2a+5b+2c}+\frac{1}{2a+2b+5c}\right)\)

\(\le3-2\left(a+b+c\right)\left(\frac{9}{9a+9b+9c}\right)=1\)

Dấu "=" xảy ra <=> a = b = c

Answer 5

Dùng Bunyakovski chú ý dấu bằng.