Problem of the week 3

[Math Tutor]

Find all natural numbers [tex]n[/tex], for which the the inequality [tex]ab^n+bc^n+ca^n+abc\le 4[/tex] is true about all positive real numbers [tex]a,b,c[/tex], for which [tex]a+b+c=3[/tex].

[dduclam]

Find all natural numbers [tex]n[/tex], for which the the inequality [tex]ab^n+bc^n+ca^n+abc\le 4[/tex] is true about all positive real numbers [tex]a,b,c[/tex], for which [tex]a+b+c=3[/tex].

The answer is [tex]n={0;1;2}[/tex]

-Case 1: [tex]n=0[/tex]. Inequality become [tex]abc\le 1[/tex] obvious (by AM-GM).

-Case 2: [tex]n=1[/tex] . Inequality become [tex]ab+bc+ca+abc\le4[/tex] also obvious (by AM-GM).

-Case 3: [tex]n=2[/tex]. Inequality become [tex]ab^2+bc^2+ca^2+abc\le4[/tex]

WOLG assume b is between a and c then [tex]a(b-a)(b-c)\le 0[/tex]

[tex]\Leftrightarrow ab^2+bc^2+ca^2+abc\le a^2b+bc^2+2abc=b(a+c)^2=\frac1{2}.2b(a+c)(a+c)\le\frac1{2}(\frac{2(a+b+c)}{3})^3=4.[/tex]

For [tex]n\ge 3[/tex],above inequality is incorrect (exam [tex]a=2,b=1,c=0[/tex]).

The Problem of the week 3 was solved.

[Math Tutor]

Excellent solution!
Thank you dduclam

[dduclam]

Thank you,techer!
We have many solutions for inequality [tex]a^2b+b^2c+c^2a+abc\le4[/tex],but above proof is nicest,i think.

Besides,we have a nother result: For [tex]a,b,c\ge0[/tex] such that [tex]a+b+c=3[/tex] and [tex]n\in N[/tex] then
[tex]a^nb+b^nc+c^na\le max[4;\frac{3^{n+1}n^n}{(n+1)^{n+1}}][/tex]

Solution for above result is not hard. And notice that above result's also true for all [tex]n>0 (n\in R+),[/tex] but in this case,we problem becom very very hard !