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let function $f(x)=ax^2+b$, find all positive real numbers $(a,b)$,such for any real numbers,then we have $$f(xy)+f(x+y)\ge f(x)f(y)$$

My try:

since $$f(xy)+f(x+y)\ge f(x)f(y)\Longrightarrow a(xy)^2+b+a(x+y)^2+b\ge (ax^2+b)(ay^2+b)$$

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Source, please? – Gerry Myerson Oct 13 '13 at 6:11
I think this problem from this picture – user94270 Oct 13 '13 at 6:15
Had you tried simplifying the inequality you derived? – Hurkyl Oct 13 '13 at 9:34
up vote 2 down vote accepted


$ax^2y^2+a(x^2+y^2)+2axy+2b\ge a^2x^2y^2+ab(x^2+y^2)+b^2$

$2b-b^2\ge (a^2-a)x^2y^2+(ab-a)(x^2+y^2)-2axy$....(*)

$\frac{2b-b^2}{a}\ge (a-1)x^2y^2+(b-1)(x^2+y^2)-2xy$

In the last part, we assumed that $a\neq0$. Since the left side is a constant, the right side must have the upper bound which is less than or equal to the left side. If $a-1> 0$, the right side doesn't have the upper bound. If $a-1<0$, $b-1>0$, and $y=0$, the right side still diverges when x-> $\infty$. For similar reason, when $a-1=0$, the right side doesn't have the upper bound. Therefore, we consider the case in which $a-1<0$ and $b-1\le0$. If $x=0$, the maximum value of the right side is $0$; hence, $2b-b^2\ge0$ and $a>0$ or $2b-b^2\le0$ and $a<0$. The first case is equivalent to $2\ge b$ and $a>0$. It implies that the given inequality works if $0< a<1$ and $b\le1$. Needless to say, the second case doesn't work.

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Nice explanation. – juantheron Oct 13 '13 at 9:17 – user94270 Oct 13 '13 at 9:28
I'm sorry, but I misunderstood that a and b were integers. Please ignore this answer and look at the link shown just above. – Math.StackExchange Oct 13 '13 at 10:02

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