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How to prove that

\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}

I tried the following:

\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}

Using Chebyshev's inequality

\begin{equation*}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\le3(\frac{1}{a}\frac{1}{a}+\frac{1}{b}\frac{1}{b}+\frac{1}{c}\frac{1}{c})\end{equation*}

from first inequality follows

\begin{equation*}\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2abc\ge a+b+c\end{equation*}

equivalent to

\begin{equation*}\frac{abc}{3(a+b+c)}\ge (\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}

and by amplifying both members by 9

\begin{equation*}\frac{3abc}{a+b+c}\ge (\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}

now using mean inequality

\begin{equation*}\sqrt[3]{abc}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\end{equation*}

the inequality in question becomes

\begin{equation*}3abc\ge (a+b+c)(\sqrt[3]{abc})^2\end{equation*}

which yields

\begin{equation*}3\sqrt[3]{abc}\ge a+b+c\end{equation*}

not what I wanted.

Alexandru
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3 Answers3

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Multiplying both sides of your inequality with $abc>0$, you get equivalently that:

$ \displaystyle (ab)^2 + (bc)^2 + (ca)^2 \geq abc (a+b+c) $

This holds by the basic inequality $ \displaystyle x^2 + y^2 + z^2 \geq xy +yz+ xz $, which holds for all $x,y,z $ real.

edit: The basic inequality holds for all real, thank's to user26486, for pointing this out.

passenger
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  • thank you, it was that simple – Alexandru Jun 02 '15 at 19:21
  • You are welcome. – passenger Jun 02 '15 at 19:23
  • The basic inequality is equivalent to $(x-y)^2+(y-z)^2+(z-x)^2\ge 0$, which is true. In fact, it holds for all real $x,y,z$, not only positive. It is also trivial by rearrangement inequality. – user26486 Jun 02 '15 at 19:31
  • @user26486: Yes you are right, it is true for all reals. I forgot it, because I had in mind just the positive numbers, for which was the inequality in the problem. – passenger Jun 02 '15 at 19:33
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    This solution also shows that the inequality holds if $abc>0$ even if $a$, $b$, and $c$ are not all positive. (e.g. If $a$ and $b$ are negative, while $c$ is positive) The solution which I posted only works if $a,b,c>0$. – Dylan Jun 02 '15 at 19:48
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The inequality which you want to prove is symmetric, so we can assume without loss of generality that $a\geq b\geq c$.

Then $ab\geq ac\geq bc$ and $\frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}$.

Thus, from the rearrangement inequality, we get that

$$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq\frac{ab}{b}+\frac{ac}{a}+\frac{bc}{c}=a+c+b$$

which is what we wanted to show.

Dylan
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1

We can do a slick AM-GM "pairwise" token that I picked up from the "Cauchy masters":

$\dfrac{ab}{c} + \dfrac{bc}{a} \geq 2\sqrt{\dfrac{ab}{c}\cdot\dfrac{bc}{a}}= 2b$, and similarly: $\dfrac{bc}{a}+\dfrac{ca}{b} \geq 2c$, and $\dfrac{ca}{b} + \dfrac{ab}{c} \geq 2a$. Add them up and divide by $2$ to get the answer.

DeepSea
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