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I was recently trying to play with mean inequalities and Jensen inequality. My question is, if the following relation holds for any positive real numbers $a$ and $b$ $$(1+a)(1+a+b)\geq\sqrt{27ab}$$ and if it does, then how to prove it. By AM-GM inequality we could obtain $1+a\geq\sqrt{4a}$ and $1+a+b\geq\sqrt[3]{27ab}$, so by putting this together we obtain$$(1+a)(1+a+b)\geq\sqrt{4a}\sqrt[3]{27ab},$$but this is slightly different from what I wanted. It's probably true that for all positive $a$ and $b$ there is $\sqrt{4a}\sqrt[3]{27ab}\geq\sqrt{27ab}$, but could there be equality?

Thanks a lot.

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Use AM-GM: $$\frac{1}{2}+\frac{1}{2}+a\ge3\sqrt[3]{\frac{a}{4}}\\ \frac{1}{2}+\frac{1}{2}+a+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}\ge6\sqrt[6]{\frac{ab^3}{108}}\\\therefore(1+a)(1+a+b)\ge\left(3\sqrt[3]{\frac{a}{4}}\right)\left(6\sqrt[6]{\frac{ab^3}{108}}\right)=\sqrt{27ab}$$

Equality holds iff $\frac{1}{2}=a=\frac{b}{3}$.

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  • $\begingroup$ That is something for I was originally looking for. Very elegant solution! Thank you. $\endgroup$ – PrimeLord May 8 '16 at 23:43
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    $\begingroup$ This is just multiplying two ordinary AM-GM inequalities, not using weights? $\endgroup$ – M.M May 9 '16 at 1:22
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The two variables function defined on $[0,+\infty)^2$ $$ f(x,y)=(1+x)(1+x+y)-\sqrt{27xy} $$ has minimum $0$ in $P_0=\left(\frac{1}{2},\frac{3}{2}\right)$.

Let us check the borders of the domain. We have $$ f(x,0)=(1+x)^2>0\\ f(0,y)=1+y>0 $$ moreover$$ \lim_{x\to+\infty}f(x,y)=+\infty\quad\forall y>0\\ \lim_{y\to+\infty}f(x,y)=+\infty\quad\forall x>0 $$ we can conclude that the inequality is strictly satisfied on the whole domain, with the exception of $P_0$, where the equal sign holds.

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  • $\begingroup$ I had never thought about it that way, thanks for interesting contribution! $\endgroup$ – PrimeLord May 8 '16 at 23:44
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It's probably true that for all positive $a$ and $b$ there is $\sqrt{4a}\sqrt[3]{27ab}\geq\sqrt{27ab}$

Since these quantities are all positive, raise both sides to the sixth power. Hence the given inequality is equivalent to $$4^3a^327^2a^2b^2\ge 27^3a^3b^3$$ which in turn is equivalent to $$4^3a^2\ge 27 b$$

Unfortunately, this need not be true.

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  • $\begingroup$ Ou, you're right, thank you for correcting my mistake. $\endgroup$ – PrimeLord May 8 '16 at 16:18
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We can begin by homogenizing the inequality to the equivalent inequality

$$(c+a)(c+a+b)\ge\sqrt{27abc^2}$$

with $a,b,c\gt0$, which can then be rewritten as

$$s(s+b)\ge\sqrt{27bc^2(s-c)}$$

with $b,s\gt0$ and $0\lt c\lt s$. It's easy to see that, for any fixed $s\gt0$, the cubic $c^2(s-c)$, which is $0$ at $c=0$ and $c=s$, has a maximum at $c={2\over3}s$, and thus

$$\sqrt{27bc^2(s-c)}\le\sqrt{27b({4\over9}s^2)({1\over3}s)}=2s\sqrt{bs}\le2s\left(b+s\over2\right)=s(b+s)$$

as desired.

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  • $\begingroup$ Is there a more general or explicit definition of "homogenize" as used here? $\endgroup$ – Ovi May 9 '16 at 6:53
  • $\begingroup$ @Ovi, all I'm doing is giving both $a$ and $b$ a denominator, $c$, then clearing it out. $\endgroup$ – Barry Cipra May 9 '16 at 11:17
  • $\begingroup$ Hm ok, i tought you just set c=1 $\endgroup$ – Ovi May 9 '16 at 15:08

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