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Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$

How can I approach this using only the AM - GM inequality? Are there any other methods that do not involve the AM-GM inequality?

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2 Answers 2

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This proof uses the rearrangement inequality in addition to AM-GM. After multiplying by $abcd$ our problem is equivalent to solving $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ Let $\{a,b,c,d\}=\{w,x,y,z\}$ with $w \ge x \ge y \ge z$. We have $$ a^2cd + ab^2d + abc^2 + bcd^2 = a(acd)+b(abd)+c(abc)+d(bcd)$$ and by the rearrangement equality $$ a(acd)+b(abd)+c(abc)+d(bcd) \le w(wxy)+x(wxz)+y(wyz)+z(xyz)$$ $$ = (wx+yz)(wy+xz)$$ By AM-GM, we get $$ (wx+yz)(wy+xz) \le \left(\frac{wx+yz+wy+xz}{2}\right)^2 = \frac{1}{4} \left((w+z)(x+y)\right)^2$$ Another AM-GM gives us $$ \frac{1}{4} ((w+z)(x+y))^2 \le \frac{1}{4} \left(\left(\frac{w+x+y+z}{2}\right)^2\right)^2 = \frac{1}{4} \left(\left(\frac{4}{2}\right)^2\right)^2 = 4$$ Thus we have $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ and furthermore, we have $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$ You might be able to tweak this to avoid the use of rearrangement, but I couldn't find a way.

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  • $\begingroup$ Brilliant! This solution covers the equality case (a, b, c, d) = (2, 0, 1, 1)! I thought only using dirty techniques would give that. $\endgroup$ Commented Aug 11, 2011 at 9:43
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    $\begingroup$ Thank you! I guess I'll keep trying at a solution using only AM-GM and post it here if I ever get it. In the meanwhile, this solution has been extremely useful! $\endgroup$
    – Paul
    Commented Aug 12, 2011 at 5:39
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The Proof below does not actually fulfill the OP's question.

I believe this problem requires two applications of the AM-GM inequality.

First, $\frac{a+b+c+d}{4} \geq (abcd)^{1/4}$. This simplifies to $1 \geq (abcd)$ which is equivalent to $\frac {4}{abcd} \geq 4$. Equality occurs at $a=b=c=d=1$.

Secondly, if we apply the inequality to the right hand side of the expected result we obtain $\frac{(a/b)+(b/c)+(c/d)+(d/a)}{4} \geq [(a/b)(b/c)(c/d)(d/a)]^{1/4}$. The terms on the right side of this inequality cancel to one and we can multiply both sides by $4$ to obtain $(a/b)+(b/c)+(c/d)+(d/a)\geq 4$. Again, equality occurs at $a=b=c=d=1$.

$\frac {abcd}{4} \leq 1$ because the product $abcd \leq 1$ . $(a/b)+(b/c)+(c/d)+(d/a) \geq 1$.

Thus, $\frac {abcd}{4} \leq (a/b)+(b/c)+(c/d)+(d/a)$

Which is not the desired result.

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    $\begingroup$ you have $a/b + b/c + c/d + d/a \geq 1$, not $a/b + b/c + c/d + d/a \leq 1$ so this can't be used in the last line $\endgroup$
    – Zarrax
    Commented Aug 9, 2011 at 3:14
  • $\begingroup$ @zarrax Thank you for catching that. I will try to fix it in the morning. $\endgroup$
    – user12998
    Commented Aug 9, 2011 at 3:59

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