# Inequality: $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$

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." • 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. Jul 21, 2021 at 14:52
• @Wolgwang Thanks, I am in the process of adding what I've done so far Jul 21, 2021 at 14:52
• OP it's best you type out your attempt using MathJax, instead of adding pictures. Thanks for adding context, though. Jul 21, 2021 at 15:00
• Why did you include part 1? Jul 21, 2021 at 15:28

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.
$$\begin{eqnarray*}\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} & \stackrel{a)}{\leq} & \frac{(a+b)^2}{4(a+b)} +\frac{(b+c)^2}{4(b+c)} + \frac{(a+c)^2}{4(a+c)} \\ & = & \frac 14(a+b + b+c + a+c) \\ & = & \frac{a+b+c}{2} \end{eqnarray*}$$
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.
• 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 :) Jul 21, 2021 at 16:41