who is the braveheart to prove the inequality...it is not hard yet.....

For real numbers x; y and z, prove the above inequality

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 8 years, 4 months ago

There should be an easier solution.

Yong See Foo - 8 years, 4 months ago

there may be but this is no less important

sayan chaudhuri - 8 years, 4 months ago

@Sayan Chaudhuri – What do you mean by no less important?

Yong See Foo - 8 years, 4 months ago

@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

sayan chaudhuri - 8 years, 3 months ago

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)

Marjorie Hoang - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$