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I am asking this question on behalf of another student, who sought my help through a school mentoring scheme, and claims that the question is similar to that found in Question $1$ of the British Maths Olympiad.

Prove that for all positive real numbers $a, b, c$:

a) $(a+b)^2 \geq 4ab$

b) $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$

The person mentioned solved a) easily but struggled with b). To solve b), the person had attempted the following:

"...putting the right hand side as a single fraction and then cross multiplying as well as cancelling terms. But it turns out to be a repeated process which makes the original question even more complicated."

enter image description here

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    $\begingroup$ Add your own work, please. What did you try? $\endgroup$
    – Ritam_Dasgupta
    Jul 21, 2021 at 14:51
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    $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – Wolgwang
    Jul 21, 2021 at 14:52
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    $\begingroup$ @Wolgwang Thanks, I am in the process of adding what I've done so far $\endgroup$
    – Hari5000
    Jul 21, 2021 at 14:52
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    $\begingroup$ OP it's best you type out your attempt using MathJax, instead of adding pictures. Thanks for adding context, though. $\endgroup$
    – Ritam_Dasgupta
    Jul 21, 2021 at 15:00
  • $\begingroup$ Why did you include part 1? $\endgroup$
    – David G. Stork
    Jul 21, 2021 at 15:28

3 Answers 3

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From AM-HM inequality, we have- $$\frac {a+b}{2}\geq \frac {2ab}{a+b}$$ Similarly write the inequality for $\{b,c\}$ and $\{c,a\}$, and add. You obtain: $$\frac {a+b+c}{2}\geq \frac {ab}{a+b}+\frac {bc}{b+c}+\frac {ac}{a+c}$$ which is the required result.

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  • $\begingroup$ Thank you for your answer $\endgroup$
    – Hari5000
    Jul 21, 2021 at 15:27
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Looking at part a), I think the intention of part b) is

\begin{eqnarray*}\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} & \stackrel{a)}{\leq} & \frac{(a+b)^2}{4(a+b)} +\frac{(b+c)^2}{4(b+c)} + \frac{(a+c)^2}{4(a+c)} \\ & = & \frac 14(a+b + b+c + a+c) \\ & = & \frac{a+b+c}{2} \end{eqnarray*}

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Note that $\dfrac{ab}{a+b}=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}$. If we did not have $\dfrac{1}{b}$, then the term would be exactly $a$. However, you are adding $\dfrac{1}{b}$ to the denominator, which is positive because $b$ is positive. Therefore $a\geq \dfrac{ab}{a+b}$. You can apply the same argument to the other three terms to get the inequality you want.

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    $\begingroup$ Your argument applies well to the initial version of the inequality, but appears to not work properly for the updated question. Would you mind revisiting? $\endgroup$
    – abiessu
    Jul 21, 2021 at 15:30
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    $\begingroup$ Honestly, I am not sure how to extend my argument to apply to the updated question. I thought of a different argument, which is showing that $\dfrac{ab}{a+b}$ has a maximum when $b=a$ at $\dfrac{a}{2}$ using derivatives. But I think the solution given by @Ritam_Dasgupta is already good enough :) $\endgroup$
    – Schach21
    Jul 21, 2021 at 16:41

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