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if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$

then :

$$(a+1)(b+1)(c+1)\ge 27.$$

My try :

$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$

$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$

then?

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Using $\bf{A.M\geq G.M}$

$$\frac{a}{2}+\frac{a}{2}+1\geq 3\left(\frac{a^2}{4}\right)^{\frac{1}{3}}$$

Similarly $$\frac{b}{2}+\frac{b}{2}+1\geq 3\left(\frac{b^2}{4}\right)^{\frac{1}{3}}$$

$$\frac{c}{2}+\frac{c}{2}+1\geq 3\left(\frac{c^2}{4}\right)^{\frac{1}{3}}$$

So $$(a+1)(b+1)(c+1)\geq 27 \left(\frac{(abc)^2}{64}\right)^{\frac{1}{3}} = 27$$

equality hold when $\displaystyle a= b=c = 2$

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We have $$(a+1)(b+1)(c+1)=9+a+b+c+ab+bc+ca,$$ since $$abc=8.$$ We get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}=2$$ and $$\frac{ab+bc+ca}{3}\geq \sqrt[3]{(abc)^2}=4$$ Multplying by $3$ and adding up, we get the statement above.

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  • $\begingroup$ nice one $(+1)$ $\endgroup$ – tired Jan 25 '17 at 19:29
  • $\begingroup$ Simplest approach :-) +unity $\endgroup$ – user399078 Feb 18 '17 at 12:22
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Here is a really hands-on way to proceed.

Suppose $b\gt c$, replace both with $B, C=\sqrt {bc}$ which keeps the product the same, but makes the two equal. Now note that $$0\lt (\sqrt b-\sqrt c)^2=b+c-2\sqrt {bc}$$ or $$B+C=2\sqrt{bc}\lt b+c$$ so the sum is reduced, and looking at the components of $(a+1)(b+1)(c+1)$ we have $B+C\lt b+c$ and $a(B+C)\lt a(b+c)$ with the other terms unchanged.

The replacement therefore reduces the product, which can be reduced, therefore, unless all the terms are equal. The minimum value therefore has $a=b=c$.

Note this includes a hidden proof of the AM/GM inequality for two variables - the AM/GM inequality is the essential idea here (or this can be seen as a property of curvature - the sum being flat and the product being curved). Any proof will also use somewhere the fact that the numbers are stated to be positive (else set one of them equal to $-1$ ...). You should reflect on how the various proofs use this fact.

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By Holder $$\prod\limits_{cyc}(a+1)\geq\left(\sqrt[3]{abc}+1\right)^3=27$$

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  • $\begingroup$ thank you .what is Holder ? $\endgroup$ – Almot1960 Jan 25 '17 at 20:13
  • $\begingroup$ @Almot1960 In our case for positive variables it the following $(a+b)(c+d)(e+f)\geq\left(\sqrt[3]{ace}+\sqrt[3]{bdf}\right)^3$. $\endgroup$ – Michael Rozenberg Jan 25 '17 at 20:23
  • $\begingroup$ How is proved؟ $(a+b)(c+d)(e+f)≥(\sqrt[3]{ace}+\sqrt[3]{b d f})^3$ $\endgroup$ – Almot1960 Jan 25 '17 at 21:27
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    $\begingroup$ @Almot1960 Holder we can prove by Jensen for $f(x)=x^k$, where $k>1$. For two series Holder is the following. Let $a_i>0$, $b_i>0$, $\alpha>0$ and $\beta>0$. Hence, we have $(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$ $\endgroup$ – Michael Rozenberg Jan 25 '17 at 21:56

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