Find the minimum value for $\dfrac{a^2+b^4+c^2}{abc}$ where $a,b,c$ belongs to the positive numbers.
I tried to solve the numerator separately then to manipulate it using $abc$ but I am not getting the desired result.
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David Krell
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Swastik Sanyal
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2you may want to detail what you mean by trying to manipulate the numerator using $abc$ – Omar Dennaoui Mar 26 '24 at 17:21
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i tried to find an inequality involving a^2+b^4+c^2 >= x times abc. then if I could do that I mamight divide it by abc to get the result . but I am completely stuck – Swastik Sanyal Mar 26 '24 at 17:24
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1I can make this as small as I want: let $a = n, b = \sqrt{n}, c = n$, then your expression is $3n^2 / n^{5/2} = 3/\sqrt{n}$. If this is homework I suspect you may have miscopied it. – Michael Lugo Mar 26 '24 at 17:25
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@MichaelLugo The OP probably means that $a,b,c\in\mathbb{N}_0$. – Vincent Batens Mar 26 '24 at 17:29
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a,b,c belongs to positive real numbers – Swastik Sanyal Mar 26 '24 at 17:32
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Does this answer your question? If $x,y,z$ are positive real number number, Then minimum value of $\frac{x^4+y^4+z^2}{xyz}$ – Aig Mar 26 '24 at 17:33
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no Aig see there are two variables with power 4 but in my question it is one. I think Sir Michael Lugo has got the answer correct. – Swastik Sanyal Mar 26 '24 at 17:39
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3@VincentBatens same logic holds for natural numbers, just replace $a=c=n^2, b=n$ and take $n\to \infty$. – Macavity Mar 26 '24 at 17:44
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@Aig my trick doesn't work in that question. – Michael Lugo Mar 26 '24 at 17:59
1 Answers
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It's evident that $$f(a,b,c):=\frac{a^2+b^4+c^2}{abc} \ge 0$$ The infimum $0$ cannot be reached. We can choose $(a,b,c) = (L^3,L,L^3)$ and make $L \to +\infty$: $$f(L^3,L,L^3) = \frac{2L^6 + L^4}{L^7}\xrightarrow{L\to +\infty} 0$$
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