It's easy to prove that \(|a| + |b| \geq |a+b|\), given that a, b are real numbers.
Apply in the problem
\(2RHS \geq |2x| + 2|y| + 2|z| = 2LHS)
(\RIghtarrow LHS \leq RHS\) (proven)

Split into a few cases. First let \(a=-x,b=-y,c=-z\). WLOG, assume \(x\geq y\geq z\). Assume all are nonnegative. Then \(x+y-z\) and \(z+x-y\) are both nonnegative. Then \(LHS=x+y+z, RHS=2x+|y+z-x|\), which makes the inequality true since \(y+z-x\leq |y+z-x|\). With this we are finished with the case all of them are negative also. Because if they are all negative, \(a,b,c\) are positive so \(|a|+|b|+|c|\leq |a+b-c|+|b+c-a|+|c+a-b|\) which is equivalent to the original inequality since \(|k|=|-k|\). Then assume only \(z\) is negative. Then \(|x+y-z|=x+y+c, |z+x-y|=x-y+c, LHS=x+y+c, RHS=2x+2c+|y-c-x|\), which makes the inequality true since \(y-c-x\leq |y-c-x|\). With this we are finished with the case only \(y,z\) negative also. Because if \(y,z\) are negative, \(b,c\) are positive with \(a\) negative or zero, so\(|a|+|b|+|c|\leq |a+b-c|+|b+c-a|+|c+a-b|\) which is equivalent to the original inequality since \(|k|=|-k|\). So we are done with all cases.

@Yong See Foo
–
i,as inequality is out of my syllabus,am not so skilled for such kind of proof of inequality....but i can understand the proof...if provided the solution....and ..so..the solution turns out to be important to me.....thanking u

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIt's easy to prove that \(|a| + |b| \geq |a+b|\), given that a, b are real numbers. Apply in the problem \(2RHS \geq |2x| + 2|y| + 2|z| = 2LHS) (\RIghtarrow LHS \leq RHS\) (proven)

Log in to reply

Split into a few cases. First let \(a=-x,b=-y,c=-z\). WLOG, assume \(x\geq y\geq z\). Assume all are nonnegative. Then \(x+y-z\) and \(z+x-y\) are both nonnegative. Then \(LHS=x+y+z, RHS=2x+|y+z-x|\), which makes the inequality true since \(y+z-x\leq |y+z-x|\). With this we are finished with the case all of them are negative also. Because if they are all negative, \(a,b,c\) are positive so \(|a|+|b|+|c|\leq |a+b-c|+|b+c-a|+|c+a-b|\) which is equivalent to the original inequality since \(|k|=|-k|\). Then assume only \(z\) is negative. Then \(|x+y-z|=x+y+c, |z+x-y|=x-y+c, LHS=x+y+c, RHS=2x+2c+|y-c-x|\), which makes the inequality true since \(y-c-x\leq |y-c-x|\). With this we are finished with the case only \(y,z\) negative also. Because if \(y,z\) are negative, \(b,c\) are positive with \(a\) negative or zero, so\(|a|+|b|+|c|\leq |a+b-c|+|b+c-a|+|c+a-b|\) which is equivalent to the original inequality since \(|k|=|-k|\). So we are done with all cases.

Log in to reply

There should be an easier solution.

Log in to reply

there may be but this is no less important

Log in to reply

Log in to reply

Log in to reply