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Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$ My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$ I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.

Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+c\ge3\sqrt[3]{abc}$$ Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)\ge63$$The last thing I tought about was that I have both $a+b+c\ge9$ and $a+b+c\ge3\sqrt[3]{abc}$ so if I somehow related them I would have $\sqrt[3]{abc} \ge 3 \rightarrow abc\ge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...

I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!

user1176409
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Spasoje Durovic
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  • I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question! – AmbretteOrrisey Dec 04 '18 at 22:54
  • @AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up) – Spasoje Durovic Dec 05 '18 at 20:05
  • @ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it! – AmbretteOrrisey Dec 05 '18 at 22:03
  • But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point. – AmbretteOrrisey Dec 05 '18 at 22:17
  • @AmbretteOrrisey I think a formal proof of a=b=c=3 being the minimal value is needed but if you could somehow show that by substituting $3+\alpha$ or $3-\alpha$ leads to an increase you would be done, long story short I think that this sort of approach altought theoretically exhaustive would be painful to face, in my humble opinion it would be hopeless but I'd be more than happy to be proven wrong :) I enjoy looking at different perspectives regarding a problem – Spasoje Durovic Dec 06 '18 at 14:18
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    And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way! – AmbretteOrrisey Dec 07 '18 at 15:40

4 Answers4

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I think you mean $$a,b,c>0$$ in this case we have $$\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ so $$a+b+c\geq 9$$ and we get also $$\frac{bc+ac+ab}{3}\geq \sqrt[3]{(abc)^2}$$ so $$abc\geq 27$$ and $$ab+ac+bc\geq 27$$ putting things together we have $$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1\geq 27+27+10=64$$

Dr. Sonnhard Graubner
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12

By the super-additivity of the geometric mean / convexity of $\log(e^x+1)$ $$ (1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3 $$ and by the GM-HM inequality $$ \sqrt[3]{abc} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=3.$$

Jack D'Aurizio
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Since $ab+bc+ca= abc$ and $$ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}\implies abc\geq 27$$ Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1\geq 55+3\sqrt[3]{27} = 64$$

nonuser
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For positive variables we need to prove that $$\ln\left((1+a)(1+b)(1+c)\right)\geq\ln64$$ or $$\sum_{cyc}(\ln(1+a)-2\ln2)\geq0$$ or $$\sum_{cyc}\left(\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\right)\geq0,$$ which is true because $$f(a)=\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\geq0$$ for all $a>0$.

Indeed, $$f'(a)=\frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!

Michael Rozenberg
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  • Can you elaborate on how you got the first expression? – YiFan Tey Dec 04 '18 at 22:53
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    @YiFan presumably by taking the $\ln(\bullet)$ of both sides using the fact that $\ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $\ln(64)$ term into three equal parts to get it into the cyclic sum, those being $\ln(64)/3 = \ln(4) = 2 \ln(2).$ – CR Drost Dec 04 '18 at 23:55
  • @CRDrost Of course. Thanks for the help! – YiFan Tey Dec 04 '18 at 23:56