Clever complex number problem ??

Check out this very clever solution although Cave's one is great as well.....

The inequality is equal to:

f(10)-f(9)+f(9)-f(8)=<

=<f(20)-f(19)+f(12)-f(11) , with f being convex and f' genuinely seria (=growing).....

I am in a hurry and cannot give more explanation but I think you understood my point.......

Another solution is to add at both sides |z+9| and square them by two....But it is a bit boring...The previous one is former.....

Oh the solution of A Cave and my first one typically are very similar although they might seem different...Mine is a general proof of the Jensen inequality with though his a consecutive use of it.....The problem was very clever generally.........

Hope I was helpful.....

I myself did not know how to solve it but my mate has done it perfectly.

Lets move z --> z+8, so that it would be more convenient.

The trick is to use the inequality |a+b| <= |a| + |b| in a form like |z+(a+b)/2| <= |z+a|/2 + |z+b|/2 several times for interim

purposes.

|z+2| + |z+3| + |z+11| <= (|z | +|z+4|)/2 + (|z| + |z+6|)/2 + (|z+10| + |z+12|)/2 <= (|z| + |z+4|)/2+ (|z | +|z+4|/2 + |z+8|/2)/2 + (|z+8|/2 + |z+12|/2 + |z+12|)/2 = |z| + 3/4*|z+4| + |z+8|/2 + 3/4*|z+12| <= |z| + 3/4*|z+4| + (|z+4|/2 + |z+12|/2)/2 + 3/4*|z+12| = |z| + |z+4| + |z+12|.

Full credits to my mate.

U should simplify it in some ranges of numbers:

Z>-8

-8>Z>-10

-10>Z>-11

-11>Z>-12

-12>Z>-19

-19>Z>-20

and see which parts r + and which r -

and u will see for all the 6 conditions the answer is correct

I`ll do one for u:

-10>Z>-11:

-(z+10)+(z+11)+(z+19)<-(z+8)+(z+12)+(z+20)

z+20<z+24 ===> 0<4 which is correct.

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