bị ẩn mất rồi nói chung là bn ấy kêu gọi \(\text{}\)

bạn ấy còn nhắn với tôi là "hi, bạn cho mk nha"

For \(\Delta ABC\) right-angled triangle at  \(A\) have \(AB=AC\), uppose in the triangle that the point M satisfies \(\widehat{MBA}=\widehat{MAC}=\widehat{MCB}\). Calculate the ratio of \(MA: MB: MC\)

 positive integers less than 10 are: {1,2,3,4,5,6,7,8,9} 
So if you pick 2 of them, you end up with a subset of integers from 12 through 98, and there are 9 x 8 = 72 possible combinations (no 10's and no multiples of 11). 

The perfect squares between 12 and 98 are 16, 25, 36, 49, 64, and 81. 

So there are 6 out of 72 possibilities, 6/72 = 1/12 = 8.3%

Given \(a,b,c\) are non-negative numbers such that \(ab+bc+ca=1\). Find the minimize value of \(P=\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\)

Given $a,b,c>0$. Prove that $\frac{a^3b}{3a+b}+\frac{b^3c}{3b+c}+\frac{c^3a}{3c+a}\ge \frac{a^2bc}{2a+b+c}+\frac{ab^2c}{a+2b+c}+\frac{abc^2}{a+b+2c}$

*)Source: https://olm.vn/hoi-dap/question/999979.html

I tried AM-GM but unsuccess