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Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.

Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.

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marked as duplicate by user147263, Did, Rory Daulton, saz, Jyrki Lahtonen Nov 23 '14 at 21:17

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    $\begingroup$ An alternate and popular form of Cauchy-Schwarz Inequality is $$\sum_{cyc}\frac{a_i^2}{b_i} \ge \frac{(\sum_{cyc} a_i)^2}{\sum_{cyc} b_i}$$ for $b_i > 0$. You may recognise this form above. $\endgroup$ – Macavity Oct 26 '14 at 16:11
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    $\begingroup$ While I agree that this is kindof an interesting inequality, what is it with asking questions about "Beautiful X"? I count twelve of your current 43 questions with "beautiful" in the title. $\endgroup$ – imallett Oct 27 '14 at 0:41
  • $\begingroup$ @GraphicsResearch idk, just that more people see them $\endgroup$ – Jackie Poehler Oct 27 '14 at 8:21
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Approach 1: Write the desired inequality as $$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9\tag{i} $$ and use the AM-GM inequality $x+y+z\geq 3(xyz)^{1/3}$ to each sum in the parentheses above.


Edit: just a few more approaches (among potentially many others). Hope you'll find this useful.

Approach 2: Multiply out the LHS of (i) as $$ 1+1+1+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right) $$ and use $x+y\geq 2\sqrt{xy}$ 3 times.

Approach 3: Use Cauchy-Schwarz like Macavity suggested below: $$ \left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left(\sqrt{\frac{1}{a}}^2+\sqrt{\frac{1}{b}}^2+\sqrt{\frac{1}{c}}^2\right)\geq(1+1+1)^2=9 $$

Approach 4: Use Jensen's inequality: the function $f(x)=\frac{1}{x}$ is convex for $x>0$ so: $$ \frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\\ \geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{3}{a+b+c} $$ from which you can rearrange to obtain your claim.

Approach 5: Please also look up the Chebyshev's sum inequality from which your inequality is an immediate consequence.

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    $\begingroup$ Or directly by Cauchy-Schwarz Inequality +1. $\endgroup$ – Macavity Oct 26 '14 at 15:15
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$$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 3\sqrt[3]{abc}\times 3\sqrt[3]{\frac{1}{abc}}=9. $$

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it is $\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ $AM-HM$

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  • $\begingroup$ I think your answer is correct and indeed very good, but you should explain it a little bit. At least mention that AM and HM stand (I assume) for "arithmetic mean" and "harmonic mean", and that you are using the general result that, given a set of positive numbers, $1/AM \leq HM$ $\endgroup$ – Luis Mendo Oct 26 '14 at 18:14
  • $\begingroup$ yes that is correct $\endgroup$ – Dr. Sonnhard Graubner Oct 26 '14 at 18:18
  • $\begingroup$ If that is correct, @Dr.SonnhardGraubner, then edit it into your answer, instead of leaving it buried in comment where no one will see it. $\endgroup$ – TRiG Oct 26 '14 at 23:30
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Divide both sides of your inequality by (a+b+c), so you now have

$$bc+ac+ab\geq\frac {9abc}{(a+b+c)}$$

Now divide both sides by a so you get

$$\frac{(bc)}{a} + c +b \ge\frac {9bc}{(a+b+c)}$$

Next divide by b

$$\frac c a + \frac a b + 1 \ge \frac {9c}{(a+b+c)}$$

Finally, divide by c

$$\frac 1 a + \frac 1 b + \frac 1 c \ge \frac 9{(a+b+c)}$$

(I have used > instead of greater than or equal to because I can't get that symbol on my keyboard.)

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  • $\begingroup$ You can type $\geq$ it as \$\geq\$. $\endgroup$ – Kim Jong Un Oct 26 '14 at 15:17
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    $\begingroup$ So you are multiplying the inequality by $abc$ and then dividing it by $abc$. Or what have you done? $\endgroup$ – Redundant Aunt Oct 26 '14 at 15:32

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