Question

Cho `a, b,c` là các số thực dương.

Tính `Max P = 4/(sqrt(a^2+b^2+c^2 + 4)) - 9/((a+b)sqrt((a+2c)(b+2c))`.

Answer 1

ngủ đi  ko ai giúp đc em đâu

Answer 2

\(\left(a+b\right)\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\dfrac{\left(a+b\right)\left(a+b+4c\right)}{2}=\dfrac{\left(a+b\right)^2}{2}+2ac+2bc\)

\(\le a^2+b^2+a^2+c^2+b^2+c^2=2\left(a^2+b^2+c^2\right)\)

\(\Rightarrow P\le\dfrac{4}{\sqrt{a^2+b^2+c^2+4}}-\dfrac{9}{2\left(a^2+b^2+c^2\right)}\)

Đặt \(\sqrt{a^2+b^2+c^2+4}=x>2\)

\(\Rightarrow P\le\dfrac{4}{x}-\dfrac{9}{2\left(x^2-4\right)}=-\dfrac{\left(x-4\right)^2\left(5x+8\right)}{8x\left(x^2-4\right)}+\dfrac{5}{8}\le\dfrac{5}{8}\)

Dấu "=" xảy ra khi \(x=4\Rightarrow a=b=c=2\)