Ta có : 

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

\(\Leftrightarrow\)\(\sqrt{A^2}+\sqrt{B^2}\ge\sqrt{\left(A+B\right)^2}\)

\(\Leftrightarrow\)\(\sqrt{A^2}+\sqrt{B^2}\ge\sqrt{A^2+2AB+B^2}\)

\(\Leftrightarrow\)\(\left(\sqrt{A^2}+\sqrt{B^2}\right)^2\ge\left(\sqrt{A^2+2AB+B^2}\right)^2\)

\(\Leftrightarrow\)\(A^2+2\sqrt{\left(AB\right)^2}+B^2\ge A^2+2AB+B^2\)

\(\Leftrightarrow\)\(2\sqrt{\left(AB\right)^2}\ge2AB\)

\(\Leftrightarrow\)\(\left|AB\right|\ge AB\) ( luôn đúng ) 

Dấu "=" xảy ra khi \(AB\ge0\)

Chúc bạn học tốt ~ 

\(\frac{c}{ab+1}\le\frac{c}{a+b}\)

\(\frac{a}{bc+1}\le\frac{a}{b+c}\)

\(\frac{b}{ac+1}\le\frac{b}{a+c}\)

\(\Rightarrow\)\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\le\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

suy ra.....

Truy cập link này nhé:

https://olm.vn/hoi-dap/question/601162.html?auto=1

CHO \(H=\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le2\)

ta có: \(0\le a\le b\le c\le1\Rightarrow abc\ge0\)

mà \(\frac{a}{bc+1}\le\frac{2a}{abc+2}\)

\(\Leftrightarrow2a+a^2bc\le2a+2abc\)

\(\Leftrightarrow abc.\left(2-a\right)\ge0\)

tương tự ta có: \(\frac{b}{1+ac}\le\frac{2b}{2+abc};\frac{c}{1+ab}\le\frac{2c}{2+abc}\)

\(\Rightarrow H\le\frac{2.\left(a+b+c\right)}{2+abc}\)

\(\Rightarrow H-2\le\frac{2.\left(a+b+c-2-abc\right)}{2+abc}=\frac{2.[\left(1-a\right).\left(1-b\right).\left(1-c\right).\left(1-ab\right)]}{2+abc}\le0\)

mà \(0\le a\le b\le c\le1\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ac+1}\le2\)  ( đ p c m)

CHÚC BN HỌC TỐT!!!!!!

giúp mik vs mik k choa

Cậu tham khảo link này , bạn chịu khó viết nha :

https://olm.vn/hoi-dap/detail/3980234685.html

Chúc bạn hok tốt

theo đề: ab = cd hay a/d = c/b 

đặt a/d = c/b = k (với k thuộc N)

=> a = kd ; c = kb

từ đó

A = (kd)ⁿ + bⁿ + (kb)ⁿ + dⁿ 

A = kⁿ(dⁿ + bⁿ) + (dⁿ + bⁿ)

A = (kⁿ + 1)(dⁿ + bⁿ)

Vậy A là hợp số \(\forall n\in N\)