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

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

Note by Sayan Chaudhuri
4 years, 11 months ago

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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 - 4 years, 11 months ago

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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 - 4 years, 11 months ago

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There should be an easier solution.

Yong See Foo - 4 years, 11 months ago

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there may be but this is no less important

Sayan Chaudhuri - 4 years, 11 months ago

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

Yong See Foo - 4 years, 11 months ago

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@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 - 4 years, 11 months ago

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