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
5 years, 4 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)

- 5 years, 4 months ago

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.

- 5 years, 4 months ago

There should be an easier solution.

- 5 years, 4 months ago

there may be but this is no less important

- 5 years, 4 months ago

What do you mean by no less important?

- 5 years, 4 months ago

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

- 5 years, 4 months ago