Question

[Lê Nhật Tân]

Help me:V

Cm bất dẳng thức

1.\(\left|a\right|+\left|b\right|\ge\left|a-b\right|\)

2.\(\left|a+b+c\right|\le\left|a\right|+\left|b\right|+\left|c\right|\)

3.\(\left|a+b\right|< \left|1+ab\right|\)

Answer 1

[kudo shinichi]

Có \(\hept{\begin{cases}\left|a\right|+\left|b\right|\ge0\\\left|a-b\right|\ge0\end{cases}}\)

\(\left|a\right|+\left|b\right|\ge\left|a-b\right|\)

\(\Leftrightarrow\left(\left|a\right|+\left|b\right|\right)^2\ge\left|a-b\right|^2\)

\(\Leftrightarrow a^2+2.\left|a\right|.\left|b\right|+b^2\ge a^2-2ab+b^2\)

\(\Leftrightarrow2.\left|a\right|.\left|b\right|\ge2ab\)( luôn đúng )

\(\Rightarrow\left|a\right|+\left|b\right|\ge\left|a-b\right|\)

                             đpcm

Answer 2

[Mất nick đau lòng con quốc quốc]

Gải sử.. 

\(1)\)\(\left|a\right|+\left|b\right|\ge\left|a-b\right|\)

\(\Leftrightarrow\)\(\left(\left|a\right|+\left|b\right|\right)^2\ge\left|a-b\right|^2\)

Có \(\left|a-b\right|^2=\left(a-b\right)^2\)

\(\Leftrightarrow\)\(a^2+2\left|ab\right|+b^2\ge a^2-2ab+b^2\)

\(\Leftrightarrow\)\(\left|ab\right|\ge-ab\) ( đúng ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(ab< 0\)

\(2)\)\(\left|a\right|+\left|b\right|+\left|c\right|\ge\left|a+b+c\right|\)

\(\Leftrightarrow\)\(\left(\left|a\right|+\left|b\right|+\left|c\right|\right)^2\ge\left|a+b+c\right|^2\)

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

\(\Leftrightarrow\)\(a^2+b^2+c^2+2\left|ab\right|+2\left|bc\right|+2\left|ca\right|\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow\)\(\left|ab\right|+\left|bc\right|+\left|ca\right|\ge ab+bc+ca\) ( đúng ) 

Dấu "=" xảy ra khi a, b, c cùng dấu ( cùng dương hoặc cùng âm ) 

\(3)\) Sai đề thì phải. Giả sử \(a=3;b=0\) thì \(\left|a+b\right|< \left|1+ab\right|\)

\(\Leftrightarrow\)\(\left|3+0\right|< \left|1+3.0\right|\)\(\Leftrightarrow\)\(3< 1\) ( ??? ) 

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