For real numbers x; y and z, prove the above inequality
4 years, 8 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)
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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.
There should be an easier solution.
there may be but this is no less important
What do you mean by no less important?
@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