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There are likely multiple ways of approaching these functional equations. Here's what i did for 1, without having to find f(1) explicitly.
Let x=0,y=0,z=0, we obtain f(0)(1−f(0)≥41, hence we can conclude that f(0)=21.
Let x=x,=0y=y,z=0, we obtain 2f(xy)+41−f(x)×21≥41, or that f(xy)≥f(x).
But since this is true for all y, with x=x,y=xy, we obtain f(y)≥f(x) for arbitrary x,y.. Hence, the only possible function is a constant function, except possibly for x=0.
Now, for any x=0, let x=y=z, and set f(x)=a. We get a−a2≥41, from which we can conclude that a=21.
Thus, the only solution is f(x)=21∀x.
Note: Yes, it is easy to find f(1) by setting x=y=z=1 to obtain f(1)=21. There are likely alternative approaches which uses this fact.
Step 2) The denominator will be in terms of
an1+bn1+cn1.Minimize this using Hölder's inequality to the form of a1+b1+c1,then again using AM≥HM write it in terms of s and substitute this minimized value in original denominator,but we should careful while substituting such inequalities.Thus,we need to check equality condition for each and every step.However,it is same for each step(a=b=c).
Finally,we arrive at the desired answer.
For question 2, the initial assumption of "ordered chain" isn't necessary, since the LHS and RHS are both independent of the order of the sequence. I wonder why that put that condition in, rather than simply saying "all positive".
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Comments
Functional Equations 1) Hints
Step 1: Find f(0).
Step 2: Show that for f(yz)≥f(y).
Functional Equations 2) Hints
Step 1: Guess what all the possible functions are.
Step 2: Show that if f(x)=x+c, then ff(x)=x+2c.
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For 1) you can find that f(x) is bounded above by some value and bounded below by some value. You need to find f(1) though.
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There are likely multiple ways of approaching these functional equations. Here's what i did for 1, without having to find f(1) explicitly.
Let x=0,y=0,z=0, we obtain f(0)(1−f(0)≥41, hence we can conclude that f(0)=21.
Let x=x,=0y=y,z=0, we obtain 2f(xy)+41−f(x)×21≥41, or that f(xy)≥f(x).
But since this is true for all y, with x=x,y=xy, we obtain f(y)≥f(x) for arbitrary x,y.. Hence, the only possible function is a constant function, except possibly for x=0.
Now, for any x=0, let x=y=z, and set f(x)=a. We get a−a2≥41, from which we can conclude that a=21.
Thus, the only solution is f(x)=21∀x.
Note: Yes, it is easy to find f(1) by setting x=y=z=1 to obtain f(1)=21. There are likely alternative approaches which uses this fact.
This isn't a full solution.
For first inequality,
Step 1)Use AM≥HM after writing b+c=2s−a
Step 2) The denominator will be in terms of an1+bn1+cn1.Minimize this using Hölder's inequality to the form of a1+b1+c1,then again using AM≥HM write it in terms of s and substitute this minimized value in original denominator,but we should careful while substituting such inequalities.Thus,we need to check equality condition for each and every step.However,it is same for each step(a=b=c). Finally,we arrive at the desired answer.
Nice set of problems.
For question 2, the initial assumption of "ordered chain" isn't necessary, since the LHS and RHS are both independent of the order of the sequence. I wonder why that put that condition in, rather than simply saying "all positive".
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I think that the condition gives a hint for Chebyshev's Inequality. Not really sure though.
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Can you give solutions especially the functional equations part,
The questions are tricky!